>. 


IN  MEMORIAM 
FLOR1AN  CAJORI 


&*-,• 


THE     ELEMENTS 


GRAPHICAL    STATICS 

AND  THEIR   APPLICATION  TO 

FRAMED    STRUCTURES, 

WITH  NUMEROUS  PRACTICAL  EXAMPLES  OF 

A 

CRANES—  BRIDGE,  ROOF  AND  SUSPENSION  TRUSSES—  BRACED 

AND  STONE  ARCHES—  PIVOT  AND  DRAW  SPANS- 

CONTINUOUS  GIRDERS,  &c., 

TOGETHER  WITH  THE  BEST 

METHODS     OF    CALCULATION, 

AND  CONTAINING  ALSO 

NEW    AND    PRACTICAL    FORMULAE 

FOR  THE 

Pivot  or  Draw  Span—  Braced  Arch—  Continuous  Girder,  Etc, 


A.  JAY   r>U    BQI^r  £,£.;,  p  Ph.D. 

PROFESSOR  OF  CIVIL  AKP.  iI1ECHA.'SITrA<iJ  ENGrV-SFHiWC.,   LETI5GH'  UNIVERSITY,   PENNA. 


"Peu  a  peu  les  connaissances  aleg6briques  d6viendront  moins  indispensables,  et  la  science,  re- 
duite  a  ce  qu'elle  doit  etre,  a  ce  qu'elle  devrait  etre  deja,  sera  ainsi  mise  a  la  port6e  de  cette  classe 
d'hommes,  qui  n'a  que  des  moments  fort  rares  a  y  consacrer."  [Poncelet-Traite  des  proprietea 
projectives  des  figures,  Paris,  1828.] 


With  an  Atlas  of  32  Plates. 


NEW  YOKK : 

JOHN   WILEY    AND    SON, 
15  ASTOR  PLACE. 

1875. 


COPYRIGHT,  1875,  BY 
JOHN    WILEY    &    SON, 


JOHN  F.  TROW  &  SON, 

PRINTERS    AND     STEREOTYPERS, 

205-213  Rast  izth  St., 

MEW   YORK. 


ELEMENTS   OF  GRAPHICAL   STATICS. 

NOTE  TO  STUDENTS  AND  TEACHERS. 

TEACHERS  examining  the  work  with  a  view  to  introduction  will  at 
once  perceive  that  it  was  not  intended  to  be  read  through  in  sequence 
by  any  class,  but  that  it  is  required  to  be  used  intelligently  with 
reference  to  the  degree  of  preparation  of  their  students,  with  a  knowl- 
edge of  the  relative  importance  and  relation  of  the  various  subjects 
treated,  and  with  a  view  to  the  end  desired.  Though  the  logical 
order  of  presentation  requires  a  certain  order  in  the  development  of 
the  subject  as  a  whole,  it  by  no  means  follows  that  this  order  should 
be  preserved  by  the  student,  or  that  he  should  even  be  acquainted 
with  the  whole.  In  fact  some  of  the  subjects  treated  of  are  best 
taken  up  by  the  student  at  a  later  period,  when  better  prepared  for 
their  comprehension,  others  are  best  omitted  at  least  in  the  first 
reading,  and  others  again  may  even  be  omitted  entirely  as  of  minor 
importance,  and  the  student  left  to  pursue  them  for  himself,  as  taste 
or  the  exigencies  of  practice  may  demand.  In  this  latter  respect,  as 
well  as  in  the  completeness  with  which  the  several  topics  are  dis- 
cussed, the  work  is  intended  to  serve  as  a  book  of  reference.  As  a 
Text  Book,  it  is  designed  to  teach  those  students  possessing  the 
knowledge  of  mathematics  and  mechanics  usual  to  the  senior  classes 
in  our  technical  schools  and  colleges,  how  to  find  the  conditions  of 
stability  in  every  kind  of  structure  of  common  occurrence  •  and  this 
not  alone  by  graphical  construction  but  also  by  calculation  as  well. 
Structures  of  less  common  occurrence  may  or  may  not  be  then  taken 
up  according  to  the  ability  of  the  class  and  the  time  at  disposal. 
With  the  above  end  in  view,  the  teacher  will  find  it  not  merely 
desirable  but  even  essential  to  depart  from  the  plan  of  the  work  as 
laid  down,  and  to  supplement  it  largely  by  various  examples  illustra- 
tive of  the  general  principles  and  bringing  out  as  clearly  and 
repeatedly  as  may  be  necessary  the  various  points  noticed  in  the 
text, 

In  the  work  four  separate  and  distinct  methods  of  solution  are 
given  for  such  structures  as  bridge  and  roof  trusses,  and  the  student 


XIV  ELEMENTS    OF     GRAPHICAL     STATICS. 

should  become  familiar  with  all.  Thus  there  are  two  methods  by  dia- 
gram, viz.,  by  resolution  of  forces  (Maxwell)  and  by  the  equilibrium 
polygon  (  Culmann) ;  and  two  corresponding  methods  by  calculation,, 
viz.,  by  composition  and  resolution  of  forces,  and  by  moments  (Hitter). 
To  give  these  in  such  manner  and  with  such  emphasis  that  the  student 
shall  be  conversant  with  all,  and  able  to  use  them  with  discrimina- 
tion where  in  any  case  they  best  apply,  we  recommend  the  following 
order  of  perusal : 

FIRST.  Chap.  I,  which  gives  the  first  method  by  diagram  (Max- 
welPs)y  and  such  of  the  Appendix  as  relates  to  this  chapter,  is  read. 
The  class  may  then  go  into  the  drawing  room,  and  under  the  super- 
vision of  the  teacher  actually  solve  a  variety  of  roof  trusses  from 
simple  to  most  complex,  both  for  dead  load  and  ivind  force  in  each 
case.  In  each  and  every  case  also  the  results  should  be  checked  by 
calculation  by  the  method  of  moments  (.Ritter's),  at  first  thoroughly 
and  in  detail,  and  afterwards  only  a  few  test  pieces  to  check  the 
accuracy  of  the  diagram.  From  Roof  Trusses  we  then  pass  on  ta 
Bridges,  and  here  also  a  series  of  selected  examples  of  every  class 
used  in  practice  is  solved,  and  the  method  of  tabulation  of  apex  weights- 
referred  to  in  Art.  12  and  Appendix  to  Chap  I,  brought  out  repeat- 
edly until  the  student  has  thoroughly  mastered  it,  and  appreciates 
fully  the  fact  that  for  each  form  of  truss  the  strains  due  to  only  two- 
weights  are  really  necessary  to  be  found,  and  that  the  others  may 
then  readily  be  found  directly  from  these.  Here  also  each  example 
should  be  checked  by  calculation  by  the  method  of  moments.  In 
the  case  of  curved  flanges  the  various  lever  arms  may  first  be 
measured  directly  to  scale  from  the  frame,  and  then  trigonometrically 
computed.  At  this  point  the  student  is  then  already  in  possession  01* 
two  independent  methods  of  solution  for  any  kind  of  framed  roof  or 
bridge  which  occurs  in  general  practice. 

There  are  in  fact  only  two  framed  structures,  the  continuous 
girder  and  the  braced  arch  remaining,  which  he  is  not  able  to  solve. 
Should  it  be  deemed  undesirable  to  consider  these,  he  can  at  once 
pass  to  Chap.  II  and  the  stone  arch.  If,  however,  a  knowledge  of 
these  is  desired,  he  is  now  ready  to  extend  his  principles  and  methods 
to  them  also.  Thus  he  has  already  recognized  that  provided  only  all 
the  outer  forces  are  known  he  can  both  diagram  and  calculate  any 
framed  structure.  In  such  structures  as  he  has  hitherto  had,  these 
outer  forces  are  either  given  or  are  easily  found.  In  the  cases  now 
considered  they  are  not  all  given,  and  must,  therefore,  first  be  found. 
Once  known,  however,  his  way  is  clear.  Kecognizing  clearly,  then, 
what  is  aimed  at,  the  supplements  to  Chaps.  VII  and  XIII  are  first 


ELEMENTS     OF     GRAPHICAL     STATICS.  XV 

taken  and  he  is  now  able  to  find  for  the  continuous  girder,  the  outer 
forces  required.  Chap.  XIII  will  then  give  exercises  in  finding  these 
forces,  and  handling  the  formula  he  has  just  deduced.  Finally,  Chap. 
XII  resumes  again  in  the  light  of  his  present  knowledge,  the  same 
old  two  methods  with  which  he  is  already  so  familiar,  of  diagram 
and  calculation,  and  a  few  examples  actually  worked  out  by  both 
methods  complete  his  mastery  of  the  continuous  girder  and  draw 
span.  He  can  now  pass  on  to  the  braced  arch,  and  in  Chap.  XIV 
will  find  all  that  he  needs.  Here  he  must  take  at  first  the  formulae 
and  constructions  for  finding  the  outer  forces,  on  trust.  Afterwards, 
if  deemed  desirable,  he  can  follow  out  the  development  of  these 
formulae  as  given  in  supplement  to  Chap.  XI Y. 

In  the  case  of  the  parabolic  arch,  at  least,  the  constructions  are  so 
simple  that  it  is  well  adapted  to  .class  instruction.  The  draw  span  is 
of  such  importance  as  to  render  some  atttention  to  it,  at  any  rate, 
desirable  in  any  full  course.  Thus  the  student  is  now  able  to  solve 
any  case  whatever  of  framed  structure  in  two  ways,  by  diagram  and 
calculation.  The  same  simple  principles  have  been  applied  throughout, 
and  formulae  have  been  called  in  only  in  a  subsidiary  way  to  deter- 
mine certain  forces  which  are  necessary  to  be  first  known  before 
these  principles  can  be  applied. 

Thus,  Chap.  I,  appendix  to  Chap.  I,  supplements  to  Chaps. 
VII  and  XIII,  then  Chaps.  XIII,  XII,  with  appendix,  and 
XI Y  with  appendix,  form  by  themselves  and  in  the  order  a 
complete  and  systematic  course. 

If,  however,  it  is  deemed  undesirable  to  consider  the  continuous 
girder  and  braced  arch,  the  course  indicated  in  the  preceding  para- 
graph may  be  omitted,  and  then  Chap.  I,  with  appendix,  constitutes 
a  course  as  thorough  as  can  be  desired,  the  student  taking  first  those 
examples  given  in  the  book,  and  then  such  others  as  the  teacher 
may  select,  and  always  solving  in  two  ways,  by  diagram  and  calcula- 
tion. He  may  then  take  Chaps.  II- Y,  and  now  possesses  a  second 
method  of  diagram  (Culmann's)  by  which  he  may  check  any  or  all  of 
the  previous  cases.  In  general,  a  few  examples  of  application  to 
bridge  girders,  drawing  the  parabolas  for  total  load  (moments)  and 
moving  load  (shear)  will  be  sufficient.  He  is  now  ready  to  pass  on 
to  the  Stone  Arch,  where  again  suitable  problems  should  be  proposed 
by  the  teacher  and  solved  under  his  supervision.  The  remainder  of 
the  work,  including  moment  of  inertia  and  continuous  girder  treated 
by  the  second  method  of  diagram,  will  in  general  be  found  unneces- 
sary and  rather  advanced,  except  in  a  very  full  course. 


Xvi  ELEMENTS     OF     GRAPHICAL     STATICS. 

The  course  recommended  is  then  as  follows,  in  the  order  given, 
chaps,  in  brackets  being  omitted  or  not  at  option  of  teacher : 

Chap.  I  and  appendix,  supplements  to  Chaps.  YII  (and  XIII, 
Chaps.  XIII,  XII  and  appendix,  XI V  and  appendix),  Chaps. 
II-Y,  XY  (XVI,  VI,  XI).  It  will  be  seen  that  the  order  in- 
tended for  class  instruction  is  quite  different  from  that  of  the 
work  itself. 

We  would  ask  teachers  examining  the  book  with  a  view  to  adop- 
tion, to  look  it  over  in  the  order  above  given. 


PREFACE. 


IT  is  now  ten  years  since  the  appearance  of  the  Graphical 
Statics  of  Culmann,*  during  which  time  the  method  has  been 
greatly  extended  in  its  applications,  and  has  met  with  such 
acceptance  that  there  is  now  scarcely  a  Polytechnikum  in  Ger- 
many where  it  is  not  a  prominent  feature  in  the  regular  course 
of  instruction. 

This  rapid  spread  of  a  new  discipline  is  the  more  remarkable 
when  we  consider  the  obstacles  which  it  encountered.  Cul- 
mann,  with  a  boldness  which  we  might  almost  term  rash,  based 
his  development  upon  the  modern  geometry  of  Yon  Staudt,  and 
assumed  in  his  re  iers  a  familiarity  with  this  very  terse^)resen- 
tation  of  a  subject  then,  as  indeed  now,  but  little  known,  and 
which,  therefore,  but  few  possessed.  To  practical  engineers, 
therefore,  to  whom  his  methods  specially  recommended  them- 
selves, his  presentation  of  those  methods  was  almost  unintelli- 
gible. 

At  a  time  when  the  students  of  the  Zurich  Polytechnic  were 
already  overburdened,  the  new  discipline  was  introduced ;  while, 
owing  to  want  of  familiarity  with  the  fundamental  principles 
premised,  they  were  unable  to  understand  his  lectures  or  read 
his  work.  Yet  such  was  the  intrinsic  value  of  the  new  method 
that,  notwithstanding  these  obstacles,  even  in  spite  of  them,  it 
made  rapid  headway  ;  found  friends  everywhere ;  crept  into 
other  departments  of  the  Polytechnic  ;  and  finally  the  aim  of 
Culmann  was  completely  attained  when  the  modern  geometry 
was  itself  introduced,  and  a  special  lecturer  in  that  branch  ap- 
pointed. Thus,  as  a  direct  result  of  the  Graphical  Statics  of 
Culmann,  appeared  the  first  and,  till  now,  only  complete  text- 
book upon  the  modern  geometry,  viz.,  Reye's  "  Geometric  der 
Lage"  Hannover,  1868.  Since  then,  hand  in  hand  and  with 
remarkable  rapidity,  these  two  studies  have  made  their  way, 

*  Die  Graphische  Statik.  Culmann.  Zurich,  1866.  Second  Edition,  1st 
vol.,  1875. 


v  PEEFACE. 

until,  as  already  remarked,  they  now  form  a  notable  feature  in 
the  course  of  every  technical  institution  in  the  land. 

The  acceptance  which  the  method  has  found  in  France,  and 
the  attention  which  it  has  there  excited,  is  sufficiently  indicated 
by  the  work  of  Levy  (La  Statique  Graphique  et  ses  Applica- 
tions, Paris,  1874),  which  contains  a  very  clear  and  elegant 
presentation  of  the  principles,  though  the  applications  are  of 
the  simplest  character,  while,  as  was  perhaps  not  unnatural  in 
the  author,  the  German  origin  of  the  system  is  very  imper- 
fectly indicated,  and  the  special  methods  of  Culmann  but  little 
more  than  hinted  at. 

In  Italy  also  the  method  has  found  an  ardent  expounder  in 
the  distinguished  mathematician  Cremona  (Le  figure  recip- 
roche  nelle  statica  grafica,  Milan,  1872),  and  to  his  efforts  and 
labors  its  introduction  and  acceptance  is  due. 

In  England,  Prof.  Clerk  Maxwell,  in  the  Trans,  of  the  Royal 
Society  of  Edinburgh,  1869-70,  has  contributed  a  paper  upon 
i%  Reciprocal  Figures,  Frames  and  Diagrams  of  Forces/'  and, 
among  others,  Jenkin,  JRanken,  Bow,  and  Unwin  have  contrib- 
uted to  the  popularity  and  spread  of  "  Maxwell's  Method." 
Maxwell  and  his  followers  give,  however,  only  the  very  simplest 
applications,  based  upon  the  resolution  and  composition  of 
forces,  such  as  will  be  found  in  our  first  chapter.  The  entire 
system  developed  by  Culmann,  the  properties  of  the  "  equilib- 
rium polygon,"  upon  which  the  fruitfulness  and  value  of  the 
graphical  statics  wholly  depend,  are  unnoticed  both  by  our 
English  and  French  authors. 

The  author  feels,  therefore,  that  no  apologies  are  needed  for 
the  present  work.  Whatever  its  shortcomings  and  defects,  he 
claims  at  least  the  honor  of  making  the  first  attempt  to  intro- 
duce among  American  Colleges  and  American  Engineers  a 
knowledge  of  a  subject  of  approved  interest  and  practical  value 
to  both,  whether  regarded  as  a  geometrical  discipline  or  as  a 
most  efficient  aid  in  investigations  of  stability.  Nor  is  he  with- 
out hope  that  the  next  ten  years  may  find  the  method  as  uni- 
versally accepted  at  home  as  now  abroad. 

The  same  difficulties  certainly  have  not  here  to  be  encoun- 
tered. The  subject  as  here  presented  requires  only  a  knowl- 
edge of  the  elements  of  geometry  as  universally  taught,  and 
can  thus  be  readily  introduced  into  our  schools  as  well  as  read 
by  those  practical  engineers  for  whose  benefit  the  method 


PREFACE.  V 

seems  so  especially  designed.  A  subject  of  such  importance, 
which  has  already  endured  successfully  so  severe  a  test,  and 
made  headway  against  such  obstacles,  we  cannot  certainly  af- 
ford any  longer  to  ignore,  and  it  is  hoped  that  the  present 
work  may  serve  to  excite  a  more  general  interest  in  the  method. 

For  the  practical  engineer,  the  importance  of  graphical 
methods  needs,  indeed,  to-day  no  demonstration.  Such  methods 
are  everywhere  in  use.  But  a  simple  and  general  system 
which  shall  include  all  special  solutions — the  development  of 
the  few  principles  upon  which  all  such  solutions  are  based,  and 
from  which  they  all  flow — is  at  least  in  this  country  unknown. 
Even  in  English  literature  there  is  to  be  found  little  more  than 
the  very  elementary  deductions  of  our  first  chapter,  so  that  it 
may  justly  be  said  that  the  entire  method  owes  its-  existence 
and  development  to  the  labors  of  German  scholars  and  the  en- 
lightened appreciation  of  German  engineers.  How  thorough 
have  been  these  labors,  how  widespread  this  appreciation,  and 
how  various  are  the  applications  of  the  method  itself,  the  reader 
may  gather  from  the  Introduction  to  this  work,  and  from  the 
appended  list  of  literature  upon  the  subject.  A  glance  at  this 
list  will  also  show  that  the  selection  of  what  was  of  most  value, 
and  the  omission  of  those  applications  of  minor  importance, 
necessary  to  bring  the  present  work  within  reasonable  limits, 
and  at  the  same  time  preserve  the  logical  unity  and  complete- 
ness of  the  whole,  was  not  the  least  difficult  portion  of  our 
task.  It  would,  indeed,  have  been  easy  to  have  given  the  work 
twice  its  present  dimensions,  though  without  a  corresponding 
increase  in  value  sufficient  to  justify  the  additional  cost.  As 
it  is,  no  application  of  real  and  practical  value  to  the  engineer 
strictly  deducible  from  the  graphical  statics  has  been  over- 
looked, and  discrimination  has  been  chiefly  exercised  in  those 
departments  where  graphical  and  analytical  processes  are  still 
of  necessity  combined.  Here  we  have  selected  only  those  cases 
where  such  union  shows  itself  most  advantageous,  and  the 
graphical  constructions  most  simplify,  illustrate,  or  interpret 
the  purely  analytical  process,  and  where  such  cases,  moreover, 
presented  a  useful,  practical,  and  not  merely  theoretical  value. 

As  to  the  plan  of  the  work,  a  word  of  explanation  is  neces- 
sary. We  have  endeavored  to  keep  always  in  view  the  re- 
quirements, of  both  students  and  practitioners,  of  technical 
schools  and  practical  engineers,  and  thus  to  combine  a  text- 


VI  PREFACE. 

book  for  school  instruction,  and  a  book  of  reference  and  manual 
for  practice  as  well.  The  attempt  is  a  difficult,  if  not  a  dan- 
gerous one,  and  one  which,  in  other  departments,  has  met  with 
more  failure  than  success.  If  we  venture  to  indulge  a  hope  that 
in  this  case  at  least  partial  success  has  been  attained,  and  that 
the  attempt  to  occupy  the  two  stools  at  once  has  not  been  dis- 
astrous, our  belief  is  due  to  the  nature  of  the  subject  itself, 
and  not  to  any  overweening  estimate  of  our  own  abilities  to 
succeed  where  so  many  have  failed.  The  subject  seems,  indeed, 
especially  suited  to  such  a  method  of  treatment.  In  fact,  no 
other  would  appear  at  this  period  to  properly  meet  the  necessi- 
ties of  the  case.  Its  geometrical  principles  are  simple,  its  ap- 
plications eminently  practical.  To  present  the  principles  alone 
would  be  to  deprive  the-  study  of  its  chief  interest  and  attrac- 
tion. To  rest  content  with  a  few  practical  applications  would 
be  to  sacrifice,  in  a  great  measure,  system  and  clearness  of  pre- 
sentation. In  the  accomplishment  of  our  double  task  we  are 
fortunate  to  have  had  at  our  disposal  such  works  as  those  of 
Bauschinger  in  the  one,  and  Culmann  in  the  other  direction. 
Our  obligations  to  both  authors  are  great,  and  are  fully  indi- 
cated in  the  text.  The  same  acknowledgment  is  due,  in  greater 
or  less  degree,  to  Mohr  and  Winkler,  Hitter  and  Reuleaux.  In 
every  case  where  such  assistance  has  been  received,  due  ac- 
knowledgment has  been  made. 

For  the  historical  and  critical  Introduction,  we  are  indebted, 
with  few  alterations,  to  the  pen  of  Weyrauch*  It  will,  we  are 
sure,  prove  of  value  to  the  student,  and  serve  to  awaken  an  in- 
terest in  those  highly  important  developments  which  geometry 
has  within  the  last  decade  undergone. 

Thus  collecting  in  a  connected  form  the  scattered  results  and 
researches  of  various  authors,  it  has  been  a  pleasurable  duty  to 
recognize  the  labors  of  those  men  who  have  chiefly  contributed 
to'  this  new  branch  of  geometrical  statics,  and  to  whom  our  own 
obligations  are  so  great.  While  thus  crediting  fully  that  which 
others  have  done,  we  have  felt  the  more  justified  in  calling  at- 
tention to  any  deviations  of  our  own.  We  have  especially 
sought  to  extend  the  application  of  the  method  by  resolution  of 
'forces  (known  best,  perhaps,  as  MaxweWs  Method ) — a  method 

*  Veber  die  gra/phische  Statik — zur  Orientirung.  Yon  Dr.  phil.  Jacob  J. 
Weyrauch,  Privat  docent  an  der  polytechnischen  schule  zu  Stuttgart.  Leipzig, 
1874. 


PREFACE.  Vll 


which  bids  fair  to  obtain  widespread  recognition,  in  direc- 
tions in  which  it  has  hitherto  been  supposed  of  little  service. 
This  often,  indeed,  by  the  aid  of  analytical  results,  we  have 
been  enabled  to  do,  and  not,  as  we  conceive,  without  a  degree 
of  success.  The  formulae  used  are  always  simple  and  of  ready 
application,  and  this  union  of  analytical  results  and  graphical 
processes  the  practical  engineer  will,  we  think,  find  of  value. 
Thus,  in  the  braced  arch  (Chap.  XIY.)  and  continuous  girder 
(Chap.  XII.)  new  constructions  will  be  found,  and  both  these 
important  and  difficult  cases  may  thus  be  solved  with  an  ease, 
completeness  and  accuracy  far :  superior  to  that  of  the  pure 
graphical  method  itself.  Those  acquainted  with  the  analytical 
investigations  of  the  " braced  arch"  as  contained  in  Oapt.  Eads> 
Report  to  the  III.  and  St.  Louis  Bridge  Co.,  May,  1868  (App.), 
will  not,  we  feel  sure,  be  slow  to  recognize  the  advantages  of 
the  present  method.  The  subject  in  its  present  state  is  thus 
fairly  brought  within  the  reach  of  the  practical  Engineer  and 
Constructor. 

To  simple  girders,  contrary  to  usually  received  opinions,  by 
the  means  of  apex  loads,  the  above  method  applies  directly,  and 
without  the  aid  of  analytical  results — a  fact  which  has  been  too 
generally  passed  over  without  sufficient  notice  by.  writers  upon 
the  subject. 

We  have  devoted  considerable  space  to  the  subject  of  the 
continuous  girder,  but  not,  we  feel  sure,  more  than  its  impor- 
tance demands.  The  subject  deserves  more  attention  at  the 
hands  of  the  practical  engineer  and  constructor  than  it  has 
hitherto  received.  That  the  present  indifference  upon  the  sub- 
ject is  due  chiefly  to  lack  of  information  can  hardly  be  doubted, 
when  the  opinion  is  current,  and  is  even  endorsed  by  those  who 
are  considered  as  authorities,  that  the  complete  solution  of  the 
/problem  is  "  probably  impossible  by  reason  of  its  complexity," 
and  "  too  complex  for  mathematical  investigation."  *  Opin- 
ions like  these  are  best  met  by  the  complete  solutions  of  par- 
ticular examples,  and  in  Chapter  XII.  will  be  found  the  com- 
plete calculation  and  tabulation  of  the  strains  in  every  piece 
due  to  every  apex  load,  for  the  central  span  of  seven  continu- 
ous successive  spans,  and,  as  far  as  any  inherent  difficulties  are 
concerned,  we  might  as  well  have  taken  50  or  100  spans. 

*  OrapJiiccd  Method  for  the  Analysis  of  Bridge  Trusses.    Greene. 


Vlll  PREFACE, 

When  engineers  shall  have  become  convinced  of  the  fact  that 
there  is  in  the  continuous  girder  a  saving  of  material  amount- 
ing usually  to  from  25  to  30  per  cent,  per  truss,  and  in  the  ex- 
treme case  even  reaching  as  high  as  50  per  cent.,  as  compared 
with  the  si^rle  girder ;  and  that  the  only  objection  which  can 
be  urged — viz.,  the  influence  of  small  variations  in  level  of  the 
supports — has,  when  properly  considered,  no  force  whatever, 
we  shall  probably  hear  less  often  of  designs  contemplating  many 
successive  and  independent  spans  of  considerable  length — such 
as,  for  instance,  for  a  bridge  over  the  Hudson  at  Poughkeep- 
sie,  consisting  of  five  separate  spans  of  525  ft.  each.  Such  a 
design  would  find  little  favor  in  France  or  Germany,  where 
continuous  girders  are  more  favorably  considered,  possibly  be- 
cause the  ability  to  calculate  them  is  less  rare,  and  reflects  in 
this  respect  little  credit  upon  the  American  profession.  About 
the  facts  in  the  case  there  can  now  be  no  dispute ;  the  subject 
has  been  too  thoroughly  investigated  to  admit  of  it,  and  we 
refer  the  reader  to  the  Appendix  for  the  results.  The  mathe- 
matician and  theoretical  engineer  have  done  their  part ;  it 
remains  for  the  practical  engineer  and  constructor  to  do  theirs. 

The  present  work  contains  the  only  complete  graphical  and 
analytical  presentation  of  this  subject  in  English  professional 
literature,  and  should  it  succeed  in  causing  a  change  of  view 
in  the  above  respect  alone,  will  not  have  been  in  vain.  In  this 
connection  the  list  of  literature  upon  the  continuous  girder 
appended  to  Chap.  XIII.  may  also  be  of  service. 

We  notice  with  pleasure  in  this  direction  the  admirable  little 
treatise  of  Clemens  Herschel,  C.E.,  upon  draw  spans.*  This 
subject  is  at  least  of  admitted  practical  value,  and  we  have 
treated  it  with  a  fullness  which,  in  our  opinion,  leaves  little  to 
be  desired.  We  have  borrowed  from  the  above  work  the  con- 
ception of  the  "  Tipper"  or  draw  with  secondary  span,  which 
is  both  new  and,  as  it  would  seem,  most  adequately  represents 
the  true  state  of  the  case,  and  alluded  to  the  idea,  also  original 
with  Mr.  Herschel,  of  weighing  off  the  reactions  at  the  supports 
of  a  continuous  girder,  instead  of  measuring  the  differences  of 
level.  In  this  case,  as  in  that  of  the  continuous  girder  gene- 
rally, we  have  clearly  brought  out  the  method  of  calculation  ~by 

*  Continuous,  Revolving  Drawbridges.  Little,  Brown  and  Company,  Bos- 
ton, 1875. 


PREFACE.  IX 

apex  weights,  and  here,  indeed,  lies  the  whole  secret  of  thorough 
practical  solution.  In  fact,  from  this  point  of  view*  the  com- 
plete solution  of  a  continuous  girder  for  any  number  of  spans, 
equal  or  unequal,  offers  no  more  essential  difficulty  than  the 
calculation  of  so  many  separate  simple  girders.  That  this  is 
not  exaggeration,  but  accurate  statement  of  fact,  a  perusal  of 
Chaps.  XII.  and  XIII.  will  suffice  to  prove. 

We  cannot  leave  this  part  of  the  subject  without  acknowledg- 
ing our  indebtedness  to  Mansfield  Merriman,  C.E.,  Assistant  in 
Engineering  in  the  Sheffield  Scientific  School  of  Yale  College, 
for  the  formulae  of  the  latter  chapter.  Mr.  Merriman  has 
done  for  the  practical  solution  of  the  continuous  girder  what 
Weyrauch  has  for  its  theoretical  discussion.  We  refer  the 
student  to  the  Supplement  to  Chap.  XIII.  for  a  specimen  of 
his  method  of  discussion. 

By  the  proper  use  of  "  indeterminate  multipliers,"  the  whole 
analytical  discussion  is  most  remarkably  simplified.  The  only 
one  of  the  many  writers  upon  the  subject  known  to  us  who 
seems  to  have  hit  upon  this  treatment  is  WinMer  (Die  Lehre 
von  der  Elasticitdt  und  festigJceit).  In  Art.  144  of  the  above 
work  he  gives  formulae  similar  to  Mr.  Merriman's  for  the 
moments  at  the  supports  of  a  continuous  girder  for  all  spans 
equal  only.  He  seems,  however,  to  have  failed  to  realize  the 
true  significance  of  the  method,  or  the  important  part  played 
by  the  Clapeyronian  numbers.  Independently  of  Winkler, 
Mr.  Merriman  has  reproduced  these  formulae  in  their  true 
light,  and  applied  the  method  to  any  lengths  and  number  of 
spans,  with  any  differences  of  level  and  any  method  of  loading. 
His  formulae  are  simple,  entirely  free,  even  in  general  form, 
from  integrals,  and  are  given  in  just  the  shape  required  in 
practice.  This  compactness  renders  it  possible  for  the  engineer 
to  enter  upon  a  couple  of  pages  of  his  note-book  all  the  for- 
mulae required  for  the  thorough  calculation  of  a  continuous 
girder  of  any  number  of  spans,  equal  or  unequal ;  and  this  cal- 
culation in  any  particular  case  proceeds  iu  a  manner  precisely 
similar  to  that  of  the  simple  girder,  directly  and  without  refer- 
ence to  authorities,  tables,  points  of  inflection,  elastic  line, 
methods  of  loading,  or  any  of  the  "  other  paraphernalia  with 
which  the  subject  is  usually  encumbered." 

It  will  be  observed  that  here  and  throughout  we  have  no- 
where left  out  of  sight  analytical  processes  or  methods.  The 


X  PREFACE. 

reader  who  considers  the  present  work  as  an  attempt  to  super- 
cede^  or  even  subordinate  analytical  investigation,  misjudges 
entirely  our  aim.  So  far  from  this,  we  indulge  the  hope  that 
its  perusal  cannot  fail  to  render  familiar  the  use  of  both 
methods,  to  bring  out  their  points  of  difference  and  relative 
advantages,  to  illustrate  the  one  by  the  other,  to  enable  the 
reader  to  check  the  results  of  the  one  by  the  other,  and  in  any 
case  apply  one  or  both,  or  a  judicious  combination  of  both,  as 
may  in  such  case  be  most  advantageous  or  desirable.  This  will 
be  especially  noticed  in  the  discussion  of  the  simple  and  con- 
tinuous girder  and  of  the  braced  arch.  (Chaps.  XII.,  XIII., 
XIY.  and  XVL,  and  Appendix.) 

As  to  the  use  of  the  work,  the  practical  engineer  will  find  in 
Chap.  I.,  and  that  portion  of  the  Appendix  relating  to  this 
chapter,  an  easy  and  simple  method  of  solution  applicable  to 
any  framed  structure  having  simple  reactions,  and  including 
thus  all  varieties  of  bridge  and  roof  trusses  of  single  span.  In 
the  Appendix  he  will  find  detailed  examples  calculated  to  illus- 
trate every  practical  point  of  importance,  and  also  a  full  expo- 
sition of  Hitters  "method  of  moments"  The  principles  of 
this  chapter  alone  will  enable  him  to  solve  readily,  both  by  cal- 
culation and  diagram,  every  case  usually  arising  in  practice. 
In  problems  involving  the  moment  of  inertia  of  areas,  in  the 
case  of  the  continuous  girder,  the  braced  arch  and  stone  arch, 
as  also  the  suspension  system,  he  will  find  Chaps.  VI.,  XII., 
XIII.,  XIV.  and  XVI.  of  value;  and  in  the  perusal  of  any  or 
all  of  these  he  will,  it  is  hoped,  find  no  trouble  by  reason  of 
logical  connection  with  preceding  principles.  They  are  in  this 
respect,  as  far  as  possible,  complete  in  themselves.  We  may 
also  call  his  attention  to  Chap.  XV._,  upon  the  stone  arch, 
though  it  is  to  be  regretted  that  the  practical  importance  of  the 
subject,  in  the  present  age  of  iron,  renders  the  ease  with  which 
it  is  graphically  treated  of  less  importance  than  formerly.  For 
his  benefit  also  frequent  practical  examples  are  given  in  detail, 
so'  that  in  all  important  applications  he  can  easily  select  a 
parallel  case,  and  follow  it  out,  step  by  step,  in  the  case  in 
hand,  without  studying  up  the  whole  process  of  development 
in  order  to  place  himself  in  a  condition  to  make  use  of  the 
methods  employed.  We  would  also  refer  him  to  THE  NEW 
METHOD  or  GRAPHICAL  STATICS  (Van  Nostrand,  1875) — a  reprint 
of  a  series  of  articles  contributed  by  the  author  to  Van  Nos- 


PREFACE.  XI 

trand's  Engineering  Magazine  during  the  present  year,  where 
lie  will  find  such  a  condensed  presentation  of  the  more  essen- 
tial principles  of  the  subject  as  will  enable  him  to  follow  the 
practical  examples  in  the  present  work  without  the  perusal  of 
the  more  lengthy  preparatory  portion  here  given. 

For  the  student  much  of  the  practical  applications  may  well 
l>e  at  first  omitted.  Notably  Chaps.  VII.-XIL,  inclusive. 
Chaps.  I.-IY.  and  XIII.-XVL  will  put  him  in  complete  pos- 
session of  the  method,  and,  moreover,  enable  him  to  solve  with 
ease  any  structure,  including  the  continuous  girder,  braced 
arch,  suspension  system,  and  stone  arch,  as  well  as  all  the  more 
ordinary  forms  of  bridge  and  roof  trusses,  cranes,  etc.  Indeed, 
if  the  first-named  structures,  which  are  of  comparatively  rare 
occurrence,  are  at  first  omitted,  Chaps.  I.-IY.  alone  will  con- 
stitute a  complete  course  upon  framed  structures  so  far  as 
usually  taught  in  our  schools  at  the  present  day.  Afterwards, 
in  practice,  and  in  the  solution  of  the  particular  problems 
treated  of,  he  will,  in  common  with  the  practical  engineer,  find 
in  the  other  portions  of  the  work  and  in  the  Appendix  just 
such  assistance  as  he  needs.  We  would  also  call  the  attention 
of  the  mathematician  more  especially  to  the  investigation  in 
Chap.  V.,  Arts.  47-51,  of  the  effects  of  a  given  recurring  system 
of  moving  loads.,  the  analytical  treatment  of  which  would  be 
almost  impracticable  by  reason  of  the  complexity  of  the  for- 
mulas obtained,  and  in  this  respect  certainly  worthless,  even  if 
possible,  but  the  geometrical  treatment  of  which  gives  rise  to 
some  of  the  most  elegant  constructions  of  the  graphical  statics; 
also  to  the  Supplements  to  Chaps.  /£IIL  and  XIV.,  in  which 
the  analytical  treatment  of  the  continuous  girder  and  braced 
arch  is  given. 

Finally,  if  our  purpose  in  writing  these  pages  is  accom- 
plished, the  principles  and  methods  here  set  forth  will  be  found 
easily  acquired,  accurate  in  their  results,  and  amply  sufficient 
for  the  ready  determination  of  the  strains  in  the  various  pieces 
of  any  framed  structure  which  the  civil  engineer  can  legiti- 
mately be  called  upon  to  design. 

With  this  much  of  introduction  and  explanation,  we  present 
our  work  to  the  engineering  profession  in  America  and  to 
American  technical  colleges,  in  the  hope  that  the  spirit  which 
has  led  to  its  production,  if  not  the  method  of  its  execution, 
may  win  for  it  a  favorable  reception. 


Xll  PREFACE. 

In  this  spirit  and  in  this  hope  we  may,  we  trust,  be  allowed 
to  appropriate  the  closing  lines  of  Culmanrfs  preface — "  Und 
nunftlhre  hin — gem  hdtte  icJi  dich  zum  Fundament  einer  auf 
wissenschaftlicherer  Basis  gegrundeten  IngenieurJciinde  ge- 
mac/itj  allein  Jcaum  darf  ich  die  Hcffnung  hegen,  so  viel  Kraft 
in  mir  zu  finden,  um  das  Game  dieses  umfangreichen  Fachen 
umzuarbeiten :  das  ist  ein  Werk,  das  mir  vor  Augen  sc/iwebt, 
wie  einer  jener  alien  mittelaUerlicJien  Dome  sick  vor  dem 
Kunstler  erhob^  der  ihn  entwarf  und  der  der  Hoffnung  sick 
nicht  hingeben  konnte,  ihnje  in  seiner  Vollendung  zu  scJiauen. 

"  Dock  es  mogen  dich  A.ndere  benutzen  und  weiter  bauen" 
nnd  was  ich  nicht  kann,  werden  nieine  Nachganger  voll- 
bringen. 

NEW  HAVEN,  April  nth,  1875. 


GElsTEEAL    CONTENTS. 


PAOB 

INTRODUCTION xxv 


PART  I.— GENERAL  PRINCIPLES. 
CHAPTER  I. 

FORCES  IN    SAME   PLANE — COMMON    POINT    OF    APPLICATION.      ILLUS- 
TRATED BY  PRACTICAL  EXAMPLES 1 

CHAPTER  II. 

FORCES  IN  SAME  PLANE — DIFFERENT  POINTS  OF  APPLICATION.      PRO- 
PERTIES OF  FORCE  AND  EQUILIBRIUM  POLYGON 16 

CHAPTER  III. 
.  CENTRE  OF  GRAVITY 29 

CHAPTER  IV. 
"MOMENT  OF  ROTATION  OF  FORCES  IN  SAME  PLANE  IN  GENERAL 33 

CHAPTER  V. 

MOMENT  OF  ROTATION  OF  PARALLEL  FORCES— PRACTICAL  APPLICA- 
TIONS        36 

CHAPTER  VI. 
MOMENT  OF  INERTIA  OF  PARALLEL  FORCES — PRACTICAL  PROBLEMS.  .      61 


PART  II.— APPLICATION  TO  BRIDGES. 

A.  THE  SIMPLE  GIRDER. 

CHAPTER  VII. 

THE   SIMPLE  GIRDER  OR   TRUSS  SUPPORTED  ONLY  AT   ENDS 84 

SUPPLEMENT  TO  CHAPTER  VII. 

THE  THEORY  OF  FLEXURE. 

CHAP.  I.  Methods  of  calculation 98 

CHAP.  II.  Principles  of  the  calculus 102 

CHAP.  III.  Theory  of  flexure ,  . .     110 


XIV  GENERAL    CONTENTS. 

B.  THE  CONTINUOUS  GIRDER  OF  CONSTANT 
CROS&SECTION. 

CHAPTER  VIII. 

PAGE 

GENERAL  PBINCIPLES 125 

CHAPTER  IX. 
LOADED  AND  UNLOADED  SPANS. 141 

CHAPTER  X. 
SPECIAL  CASES  OF  LOADING 147 

CHAPTER  XI. 
METHODS  OP  LOADING  CAUSING  MAXIMUM  STRAINS 155 

CHAPTER  XII. 
COMBINATION  OP  GRAPHICAL  AND  ANALYTICAL  METHODS 168 

CHAPTER  XIII. 
ANALYTICAL  FORMULAE 198 

SUPPLEMENT  TO  CHAPTER  ,XIIL 
DEMONSTRATION  OF  ANALYTICAL  FORMULAE.  . .  239 


PAET  III.— APPLICATION  TO  THE  AKCH. 
CHAPTER  XIV. 

THE  BRACED  ARCH.      COMBINATION  OF  GRAPHICAL  AND  ANALYTICAL 

METHODS 251 

SUPPLEMENT  TO  CHAPTER  XIV. 
DEMONSTRATION  OF  ANALYTICAL  FORMULAE 271 

CHAP.  I.  General  considerations  and  formulae 271 

CHAP.  II.  Hinged  arch  in  general 277 

CHAP.  III.  Arch  hinged  at  abutments  only 283 

CHAP.  IV.  Arch  fixed  at  ends— no  hinges 287 

CHAP.  V.  Influence  of  temperature 299 

CHAP.  VI.  Partial  uniform  loading 304 

CHAPTER  XV. 

THE   STONE  ARCH 31 1 

CHAPTER  XVI. 

THE  INVERTED  ARCH— SUSPENSION  SYSTEM.  .  327 


APPENDIX..  339 


TABLE    OF    CONTENTS. 


INTRODUCTION. 

ABT.  PAGE 

I.  Upon  mathematical  investigations  generally xxvi 

II.  Analytical  and  geometrical  mechanics xxvii 

III.  Geometrical  statics xxix 

IV.  The  graphical  calculus xxxi 

V.  Graphical  representation xxxiv 

VI.  Graphical  statics xxxv 

VII.  The  methods  and  limits  of  the  graphical  statics xxxvii 

VIII.  The  modern  geometry xxxix 

IX.  The  modern  geometry  in  engineering  practice xlii 

X.  Practical  significance  of  the  graphical  statics xliii 

XI.  Literature  upon  the  graphical  statics xlv 

XII.  Graphical  dynamics 


PAET  I.— GENERAL  PRINCIPLES. 

CHAPTER  I. 

FORCES  IN  SAME  PLANE— COMMON  POINT  OF  APPLICATION. 

1.  Notation — Representation  of  forces  by  lines 1 

2.  Resultant  of  two  forces 2 

3.  Resultant  of  any  number  of  forces 2 

4.  Conditions  of  equilibrium 3 

5.  Properties  of  force  polygon 3 

6.  Order  of  forces  in  force  polygon  a  matter  of  indifference 4 

7.  Forces  acting  in  same  straight  line 5 

8.  PRACTICAL  APPLICATIONS 5 

9.  Braced  semi-arch 6 

10.  Roof  truss 8 

11 .  Diagram  for  wind  force 8 

12.  Application  of  method  to  bridges 11 

13.  Braced  arch 12 

14.  Ritter's  u  Method  of  Sections  " 14 

CHAPTER  II. 
FORCES  IN  SAME  PLANE— DIFFERENT  POINTS  OF  APPLICATION. 

16.  Resultant  of  two  forces 16 

17.  Case  of  forces  parallel 18 


XVI  TABLE    OF   CONTENTS/ 

AKT.  PAGE 

18.  Perpendiculars  let  fall  upon  components  from  intersection  of  outer 

polygon  sides,  are  inversely  as  the  components 18 

19.  Equilibrium  polygon 19 

20.  Case  of  a  couple — Conditions  of  equilibrium 20 

21.  Properties  of  a  couple 21 

22.  Force  and  equilibrium  polygons  for  any  number  of  forces 22 

23.  Influence  of  a  couple 24 

24.  Order  of  forces  in  force  polygon,  a  matter  of  indifference 25 

25.  Pole  upon  the  closing  line — Failing  case  for  equilibrium.. 25 

26.  Relation  between  two  equilibrium  polygons  with  different  poles  ....  26 

27.  Mean  polygon  of  equilibrium 27 

28.  Line  of  pressures  in  an  arch 28 

CHAPTER  III. 

CENTRE  OF  GRAVITY. 

30.  General  method  for  determination  of  centre  of  gravity 29 

31.  Reduction  of  areas 30 

32.  Reduction  of  triangle  to  equivalent  rectangle  of  given  base 31 

33.  Reduction  of  trapezoid  to  equivalent  rectangle 31 

34.  Reduction  of  quadrilateral  generally 31 

CHAPTER  IT. 

MOMENT  OF  ROTATION  OF  FORCES  IN  THE  SAME  PLANE 
IN  GENERAL. 

35.  Definition  of  moment  of  rotation 33 

36.  Culmann's  principle 33 

37.  Application  to  equilibrium  polygon 34 

CHAPTER  Y. 
MOMENT  OF  RUPTURE— PARALLEL  FORCES. 

38.  Equilibrium  polygon — Ordinates  to  give  the  moments  of  rupture.    . .  36 

39.  Beam  with  two  equal  and  opposite  forces  beyond  the  supports 38 

40.  Beam  with  two  equal  and  opposite  forces  between  the  supports 39 

41.  Special  cases  of  importance 41 

1st.  Beam — load  inclined  to  axis. ...    41 

2d.  Force  parallel  to  axis 42 

3d.  Forces,  in  different  planes '  42 

4th.  Combined  twisting  and  bending  moments 43 

5th.  Application  to  crank  and  axle 44 

42.  Loading  continuous — Load  area 45 

43.  Beam  uniformly  loaded 46 

44.  Moment  curve  a  parabola 46 

45.  Beam  continuously  loaded  and  also  subjected  to  action  of  concen- 

trated loads 48 

46.  Influence  of  moving  load 50 

47.  Load  systems 53 


TABLE    OF    CONTENTS.  XV11 

ABT.  PAGE 

48.  Properties  of  the  parabola  included  by  the  closing  line 54 

49.  Application  of  the  above  principles 54 

50.  Most  unfavorable  position  of  load  system  upon   a  span  of  given 

length 56 

51.  Greatest  moment  at  a  given  cross-section 59 

CHAPTER  VI. 
MOMENT  OF  INERTIA  OF  PARALLEL  FORCES. 

52.  Application  of  moment  of  inertia  in  proportioning  any  cross-section.  61 

53.  Graphical  determination  of  moment  of  inertia 62 

54.  Signification  of  the  area  of  the  equilibrium  polygon 63 

55.  Radius  of  gyration 63 

56.  Inertia  curves— Ellipse  and  hyperbola  of  inertia 66 

57.  Construction  of  curve  of  inertia 69 

58.  Example 69 

59.  Central  curve— Central  ellipse 71 

60.  Centre  of  action  of  the  statical  moments  considered  as  forces 73 

61.  Cases  where  the  direction  of  the  conjugate  axes  of  the  inertia  curve 

can  be  at  once  determined 75 

62.  Practical  applications 76 

1st.  The  parallelogram 76 

2d.  The  triangle 77 

3d.  The  trapezoid 78 

4th.  The  parabolic  segment 80 

C3.  Compound  or  irregular  areas 81 

64.  Cases  where  there  is  no  axis  of  symmetry 83 


PAET  II.— APPLICATION  TO  BRIDGES. 
A.  THE  SIMPLE  GIRDER. 

CHAPTER  VII. 

THE  SIMPLE  GIRDER. 

66.  Forces  which  act  upon  a  bridge 84 

67.  Bridge  loading 85 

68.  Shearing  force — Moment  of  rupture,  etc 85 

69.  Concentrated  loads — Invariable  in  position 87 

70.  Concentrated  loads— Variable  in  position 88 

71.  Position  of  a  given  system  of  loads  causing  maximum  shear. 88 

72.  Construction  of  the  maximum  shear 89 

73.  Maximum  moments 90 

74.  Construction  of  maximum  moments 91 

75.  Absolute  maximum  of  moments 92 

76.  Continuous  loading 93 

77.  Total  uniform  load 93 

78.  Method  of  loading  causing  maximum  shear 94 

79.  Live  and  dead  loads. .  95 


XV111  TABLE    OF   CONTENTS. 


SUPPLEMENT  TO  CHAPTER  VII. 
CHAPTER  I. 

METHODS  OP   CALCULATION. 

ABT.  PAGE 

2.  Hitter's  method 99 

3.  Method  by  resolution  of  forces 100 

CHAPTER  II. 

PRINCIPLES  OP   THE   CALCULUS. 

4.  Differentiation  and  integration 102 

5.  Powers  of  a  single  variable 105 

6.  Other  principles 106 

7.  Illustrations 107 

a  First  differential  coefficient 109 

CHAPTER  III. 

THEORY  OF  FLEXURE. 

9.  Coefficient  of  elasticity 110 

10.  Moment  of  inertia 110 

11.  Change  of  shape  of  axis 112 

12.  Beam  fixed  at  one  end,  load  at  other 113 

13.  Beam  as  above — Uniform  load 117 

14.  Beam  supported  at  both  ends — Concentrated  load 118 

15.  Beam  as  above — Uniform  load 119 

16.  Beam  fixed  at  one  end,  supported  at  other — Concentrated  load 120 

17.  Beam  as  above — Uniform  load 121 

18.  Beam  fixed  at  both  ends— Concentrated  load 122 

19.  Inflection  points,  etc. 124 

B.  THE  CONTINUOUS  GIRDER  OP  CONSTANT 
CROSS-SECTION. 

CHAPTER  VIII. 
GENERAL  PRINCIPLES. 

80.  Mohr's  principle 125 

81.  Determination  of  tangents  to  the  elastic  curve 127 

82.  Effect  of  the  moments  at  the  supports 128 

83.  Division  of  the  moment  area 129 

84.  Properties  of  the  equilibrium  polygon 130 

85.  Polygon  for  the  positive  moment  areas 131 

86.  Construction  of  the  fixed  points,  and  of  the  equilibrium  polygon  ...  132 

87.  Construction  of  the  moments  at  the  supports 135 

88.  The  second  equilibrium  polygon 136 

89.  Determination  of  moments  at  the  supports.. 138 

90.  Comparison  with  girder  fixed  horizontally  at  both  ends 139 


TABLE    OF   CONTENTS.  XIX 

CHAPTER  IX. 
LOADED  AND  UNLOADED  SPANS. 

ABT.  PAGE 

91.  Unloaded  span 141 

92.  Two  successive  unloaded  spans , 141 

93.  The  fixed  points 142 

94.  Shearing-  force — Reactions  at  the  supports,  and  moments  in  the  un- 

loaded spans 143 

95.  Loaded  spans 143 

96.  Two  successive  loaded  spans 144 

97.  Arbitrary  loading 145 

CHAPTER  X. 

SPECIAL  CASES  OF  LOADING. 

98.  Total  uniform  load 147 

99.  Practical  examples — Girder  of  four  spans 148 

100.  Partial  uniform  load 149 

101.  Concentrated  load 153 

CHAPTER  XI. 

METHODS  OF  LOADING  CAUSING  MAXIMUM  STRAINS. 

102.  Maximum  shearing  force 155 

103.  Maximum  moments 157 

104.  Determination  of  the  maximum  shearing  forces 159 

105.  Determination  of  the  maximum  moments 161 

106.  Practical  simplifications  of  the  method 162 

107.  Approximate  practical  constructions 163 

1st.  Beam  of  two  spans— Moments 163 

Beam  of  two  spans — Shearing  forces 164 

2d.  Beam  of  three  or  more  spans — Moments 164 

Beam  of  three  or  more  spans — Shearing  forces 165 

108.  Method  by  resolution  of  forces — Draw  spans 166 

CHAPTER  XII. 

CONTINUOUS  GIRDER  CONTINUED— COMBINATION  OF 
GRAPHICAL  AND  ANALYTICAL  METHODS. 

111.  Method   of  finding  shearing  forces,   when  inflection    points  are 

known 168 

1st.  Loaded  span 168 

2d.  Unloaded  span 169 

112.  Determination  of  inflection  points—  Inflection  verticals 170 

113.  Beam  fixed  horizontally  at  ends 171 

114.  Example 172 

1 15.  Counterbracing 174 


XX  TABLE   OF   CONTENTS. 

ART.  PAGR 

116.  Beam  fixed  at  one  end,  supported  at  the  other 175 

117.  Practical  construction 178 

118.  Beam  continuous  over  three  level  supports 178 

119.  Practical  construction 170 

120.  The  pivot  draw  with  secondary  central  span 181 

121.  Supports  not  on  a  level — Reactions 182 

122.  Beam  over  four  level  supports 184 

123.  Practical  construction 186 

124.  Pivot  span— Example 187 

125.  Method  of  passing  in  construction,  from  one  span  to  another 190 

126.  Method  of  procedure  for  any  number  of  spans 192 

127.  Example — Central  span  of  seven  spans 193 

128.  Method  of  calculation  by  moments 196 


CHAPTER  XIII. 

ANALYTICAL  FORMULA  FOR  THE  SOLUTION  OF  CON- 
TINUOUS GIRDERS. 

129.  Introduction 198 

130.  Notation 200 

131.  Theorem  of  three  moments 201 

132.  Example— Total  uniform  load— Moments 202 

133.  Triangle  of  moments 203 

134.  Total  uniform  load — All  spans  equal — Reactions 205 

135.  Triangle  for  reactions 205 

136.  Clapeyron's  numbers 206 

137.  Uniform  live  load  over  any  single  span — Moments  at  supports 207 

138.  Triangle  for  moments — Loaded  span 208 

139.  Unloaded  spans — Moments 209 

140.  Practical  rule  and  table  for  finding  the  above 210 

141.  Reactions  at  supports — Loaded  span 212 

142.  Triangle  for  reactions 213 

143.  Reactions  for  unloaded  spans — Tables  for 214 

144.  Concentrated  load  in  any  span— Moments 216 

145.  Application  of  above  formulae 216 

1 46.  Triangle  and  table  for  moments 218 

1 47.  Reactions  at  supports — Concentrated  load , 219 

148.  Shear  at  supports 219 

149.  Recapitulation  of  above  formulae 221 

150.  Continuous  girder  with  variable  end  spans 224 

151.  Application  of  formulae  for 225 

152.  Continuous  girder  with  fastened  ends 22t'> 

153.  Beam  of  single  span — Fastened  at  both  ends — Fastened  at  one  end, 

etc. — Examples 22f> 

154.  Tables  for  moments — End  spans  variable 229 

155.  Continuous  girder — All  spans  different— General  formulae — Exam- 

ples   233 

156.  General  method  of  calculation  . .  237 


TABLE    OF    CONTENTS.  XXI 

SUPPLEMENT  TO  CHAPTER  XIII. 

DEMONSTRATION  OF  ANALYTICAL  FORMULA  FOR  THE 
CONTINUOUS  GIRDER. 

IRT.  PAGfi 

1.  Conditions  of  equilibrium 239 

2.  Equation  of  elastic  line 240 

3.  Theorem  of  three  moments 241 

4.  Determination  of  moments — Supports  all  on  level £42 

5.  Uniform  load w 244 

0.  Formulae  for  the  "  Tipper  "  [Art.  120] 244 

7.  Example  of  two  span  tipper 245 

8.  LITERATURE  UPON  THE  CONTINUOUS  GIRDER.  .  247 


PART  III.— APPLICATION  OF  THE  GRAPHICAL 
METHOD  TO  THE  ARCH. 

CHAPTER  XIV. 

THE  BRACED  ARCH. 

157.  Different  kinds  of  braced  arch 251 

158.  Arch  hinged  at  both  crown  and  abutments 251 

159.  Hinged  at  abutments  only— Continuous  at  crown 253 

1st.  Parabolic  arch 254 

2d.    Circular  arch,  Tables  for  solution  of 255 

160.  Arch  fixed  at  abutments — Continuous  at  crown 257 

1st.  Parabolic  arch 258 

2d.    Circular  arch 260 

162.  General  method  of  solution 262 

163.  Analytical  formulae  for  horizontal  thrust,  and  vertical  reactions 264 

164.  Arches  with  solid  web 266 

165.  Strains  due  to  temperature,  Formulae  for 267 

166.  Effects  of  temperature 269 

SUPPLEMENT  TO  CHAPTER  XIV. 

DEMONSTRATION  OF  ANALYTICAL  FORMULA  FOR  THE 
BRACED  ARCH. 

CHAPTER  I. 

GENERAL,   CONSIDERATIONS  AND   FORMULAE. 

1.  Fundamental  equations ; 271 

2.  Displacement  of  any  point 275 

CHAPTER  II. 

HINGED  ARCH  IN  GENERAL. 

3.  Notation — The  outer  forces  in  general 277 

4.  Intersection  line 278 

5.  Parabolic  arch — Concentrated  load 278 

6.  Circular  arch — Concentrated  load 280 

7.  Integrals  used  in  above  discussion 282 


XX11  TABLE    OF   CONTENTS. 


CHAPTER  III. 
ARCH  HINGED  AT  ABUTMENTS  ONLY. 

A.  Parabolic  arc. 

ART.  PAGK 

8.  Horizontal  thrust 283 

9.  Intersection  curve 283 

B.  Circular  arc. 

10.-  Horizontal  thrust 283 

11.  Intersection  curve 285 

CHAPTER  IV. 

ARCH  FIXED  AT   ENDS. 

12.  Introduction 287 

13.  Concentrated  load — General  formulae 287 

A.  Parabolic  arc. 

14.  Determination  of  H,  V  and  M0 289 

15.  Intersection  curve 291 

16.  Direction  curve 291 

B.  Circular  arc. 

17.  Fundamental  equations 291 

18.  Determination  of  H,  V  and  M0 293 

19.  Intersection  curve 297 

20.  Direction  segments 297 

20  (b).  Transformation  series 297 

CHAPTER  V. 

INFLUENCE  OF  TEMPERATURE. 

21.  General  considerations 299 

22.  Influence  of  temperature  on  the  arch 299 

23.  Fundamental  equations — General 300 

24.  Arch  with  three  hinges 301 

25.  Arch  hinged  at  ends 501 

26.  Arch  without  hinges 301 

CHAPTER  VI. 

PARTIAL  UNIFORM  LOADING. 

27.  Notation 304 

A.  Arch  hinged  at  crown  and  ends. 

28.  Vertical  reaction 304 

39.  Horizontal  thrust 305 

B.  Arch  hinged  at  ends  only. 

30.  Vertical  reaction 305 

31.  Horizontal  thrust— Parabolic  arch 305 

32.  Horizontal  thrust— Circular  arch 306 

O.  Arch  without  hinges — continuous  at  crown. 

33.  Parabolic  arch.     Formulae  for  V,  H  and  M 307 

34.  Circular  arch.     Formulae  for  V,  H  and  M 308 


TABLE    OF   CONTENTS.  XX111 

CHAPTER  XY. 

THE  STONE  ARCH. 

ART.  PAGB 

167.  Definitions,  etc ; 

168.  Pressure  line 311 

169 .  Sliding  of  the  joints 311 

170.  Forces  acting  upon  a  cross-section — Neutral  axis 312 

171.  Kernel  of  a  cross-section 313 

172.  Position  of  kernel  for  different  cross-sections 314 

173.  Proper  position  of  resultant  pressure 316 

174.  Pressure  line — True  pressure  line 317 

175.  Support  line   318 

176.  Deviation  of  support  from  pressure  line 319 

177.  Dimensions  of  the  arch 320 

178.  Construction  of  the  pressure  line 322 

179.  Practical  example " 324 

180.  Proper  depth  of  arch  at  crown 324 

181.  Increase  of  depth  due  to  change  of  form 326 

CHAPTER  XYI. 
THE  INVERTED  ARCH— SUSPENSION  SYSTEM. 

183.  Methods  of  construction 327 

184.  Rear  chains,  and  anchorages 327 

185.  Cable  with  auxiliary  stiffening  truss 329 

186.  Method  of  loading  causing  maximum  strains 330 

187.  Example ,, 333 

188.  Analytical  investigation  of  the  forces  acting  upon  the  stiffening 

ooo 

truss 066 

189.  Concluding  remarks 335 


APPENDIX. 

NOTE  TO  CHAP.  VIII.  OP  THE   INTRODUCTION— UPON  THE  MODERN 
GEOMETRY. 

NOTE  TO  CHAPTER  1 346 

2.  Bent  crane 346 

3.  Character  of  strains  in  the  pieces  as  indicated  by  the  strain  diagram,  347 

4.  Pieces  in  equilibrium — Points  to  be  avoided  in  constructing  the 

strain  diagram 347 

5.  Roof  truss — Method  of  checking  the  accuracy  of  the  diagram 348 

6.  The  French  roof  truss — Solution  apparently  indeterminate — Method 

of  solution — Method  of  calculation  by  moments 348 

7.  Application  to  bridges — Bowstring  girder — Method  of    tabulation 

illustrated. 350 

8.  Strains  in  the  flanges— Table 353 

9.  Method  of  calculation  by  moments 354 

10.  Girder  with  straight  flanges — Howe  or  Murphy  Whipple  system  of 

.      bracing 355 


XXIV  TABLE    OF    CONTENTS. 

ART.  ,PAGB 

11.  Lenticular  girder,  or  system  of  Pauli 358 

12.  Remarks  upon  above  system — The  most  economical  system  for  long 

spans 359 

NOTE  TO  CHAPTER  II. 

14.  Equilibrium  polygon  considered  as  a,  frame 362 

NOTE  TO  CHAPTER  V.,  ART.  51. 

15.  Beam  subjected  to  given  force  system — Maximum  moment  at  any 

cross-section 364 

NOTE  TO  CHAPTER  XII. 

16.  Pivot  or  draw  span — Practical  example — Solution  by  diagram — By 

method  of  moments — By  resolution  and  composition  of  forces ....     365 

17.  Relative  economy  of  the  continuous  girder 374 

18.  Continuous  girder — supports  out  of  level 377 

NOTE  TO  CHAPTER  XIV.  THE  BRACED  ARCH. 

19.  Practical  example  of  the  braced  arch 382 

20.  Arch  hinged  at  both  abutments  and  crown 383 

21.  Arch  hinged  at  abutments  only 386 

23.  Temperature  strains  for  above 391 

26.  Arch  continuous  at  crown — Fixed  at  ends 394 

27.  Temperature  strains  for  above 398 

28.  Advantage  of  arch  with  fixed  ends  for%long  spans — Comparison  with 

the  St.  Louis  arch. .  399 


INTRODUCTION. 


HISTORICAL   AND    CRITICAL.*  I  ^/t?     I 

X^  '  ° 

THE  subject  of  Graphical  Statics  has,  since  the  appearance  of  (/ulrnann's 
work  (Die  grapliisclie  Statik,  Zurich,  Meyer  and  Zeller,  18#t>),  excited 
considerable  attention,  but  an  accurate  and  just  estimate  of  its  methods 
and  practical  value  is  still  wanting.  Thus  there  are  some  who  oppose  it; 
others  willingly  accept  it  as  an  efficient  and  valuable  aid  in  practical  inves- 
tigations of  stability ;  still  others  even  profess  to  see  in  it  a  future  rival  of 
Analytical  Statics.  This  last  somewhat  remarkable  claim  seems  apparently 
justified  by  a  passage  in  Culniann's  preface,  where  it  is  asserted  "  that  the 
Graphical  Statics  will  and  must  extend,  as  graphical  methods  find  ever 
wider  acceptance — but  in  such  case,  however,  its  treatment  will  soon  escape 
the  hands  of  the  practitioner,  and  it  will  then  be  built  up  by  the  geometer 
and  mechanic  to  a  symmetrical  whole,  which  shall  hold  the  same  relation 
to  the  new  geometry  that  analytical  mechanics  does  to  the  higher  analysis." 
These  various  and  conflicting  opinions  find  their  supporters  in  technical 
schools  and  among  engineers  throughout  Germany. 

In  the  consideration  of  the  subject,  we  shall  endeavor  especially  to  give 
an  objective  presentation,  but  shall  also  feel  at  liberty  to  present  our  own 
opinions  as  well,  and  generally  to  venture  such  reflections  as  seem  suited 
to  throw  light  upon  the  matter.  For  both  reasons  it  will  sometimes  be 
necessary  to  make  apparent  deviations,  in  order  to  point  out  the  various 
fields  in  which  these  new  investigations  take  root,  to  define  their  limits,  and 
to  decide  in  what  directions  and  to  what  extent  impulse  and  sustenance 
for  further  development  may  exist.  In  such  a  manner  only  can  we  satis- 
factorily ascertain  how  far  the  graphical  statics  may  safely  count  upon 
more  than  a  passing  recognition  and  brief  existence. 

We  have  therefore  to  ask  of  the  reader  who  wishes  to  obtain  a  just  and 
accurate  estimate  of  this  new  and,  as  we  venture  to  think,  highly  important 
subject,  patience  for  the  following  general  considerations. 

*  Ueber  die  graphische  Statik— zur  Orientirung.  By  J.  I.Weyrauch.  Leip- 
zig, 1874. 


XX  Vl  MATHEMATICAL   INVESTIGATIONS.  [iNTIiOD. 


L 

y 

UPON   MATHEMATICAL   INVESTIGATIONS    IN    GENERAL. 

Mathematical  truths  may  be  attained  in  two  essentially  different  methods 
— by  synthesis  or  by  analysis,  by  composition  or  by  resolution.  In  synthe- 
sis, we  ascend  from  particular  cases  to  general  ones ;  in  analysis,  we  descend 
from  general  cases  to  particulars.  By  synthesis  we  pass  from  the  simplest 
or  admitted  truths,  by  combination  and  comparison,  to  more  complicated 
phenomena.  Analysis  seeks  to  refer  back  such  phenomena  to  their  fun- 
damental relations,  or  to  deduce  special  properties  from  the  general  con- 
ditions. 

The  analysis  of  a  phenomena  presupposes,  then,  an  accurate  comprehen- 
sion of  all  its  elements.  So  far  as  these  last  stand  in  relations  of  cause 
and  effect  to  the  whole  and  its  parts,  or  so  far  as  such  relations  exist  be- 
tween the  parts  themselves,  they  may  be  expressed  by  equations.  Thus 
the  operations  which  are  necessary  in  analysis  become  independent  of  con- 
crete phenomena,  and  are  governed  only  by  the  laws  of  abstract  quantities 
as  included  by  algebra  in  the  widest  sense  of  the  word.  Algebra,  then,  is 
not  analysis  itself,  but  only  its  instrument,  "  instrument  precieux  et  neces- 
saire  sans  doute,  parce  qu'il  assure  et  facilite  notre  marche,"mais  qui  ri>a  par 
lui  meme  aucune  vertu  propre  ;  qui  ne  dirige  point  I' *  esprit,  mais  que  V  esprit 
doit  diriger  comme  tout  autre  instrument'1''  (Poinsot,  Theorie  nouvelle  de  la 
rotation,  pre"s  a"  T  Acad.,  1834).  Ordinarily  the  higher  branches  of  algebra, 
with  which  numberless  really  analytical  investigations  are  connected,  are 
designated  as  analysis.  More  properly,  all  investigations  which  rest  upon 
equations  of  condition  may  be  termed  analytical  investigations. 

Synthetic  investigation  rests  mainly  upon  geometrical  conceptions,  and 
attains  to  the  knowledge  of  phenomena  through  concrete  conditions,  which 
latter  may  be  designated  as  space  relations  and  processes.  Hence  the  usual 
division  into  analytical  and  geometrical  methods,  even  in  applied  mathe- 
matics. We  have  thus  with  equal  appropriateness  an  analytical  geometry 
as  also  a  geometrical  analysis.  When  pure  geometry  (in  distinction  from 
analytical)  makes  use  of  the  symbols  and  operations  of  algebra,  it  is  only 
to  express  with  corresponding  generality  and  more  concisely  than  in  words 
truths  attained  to  by  abstraction,  and  independent  of  the  dimensions  of  the 
auxiliary  figure ;  or  so  to  formulate  such  truths  that  they  may  be  applied 
in  analytical  investigation.  Accordingly,  such  use  of  algebraic  formulae 
has  as  little  effect  upon  the  synthetic  process  as  from  the  above  it  would 
seem  essential  to  the  analytic  treatment.  In  either  case,  algebra  is  but  the 
instrument,  the  method  lies  back  of  and  directs  it. 

If  analytical  formulae  and  operations  are  entirely  excluded  from  the 
more  complicated  geometrical  investigations,  we  are  at  once  restricted  to 
general  laws  of  metrical  relation.  There  remains  only  the  faculty  of 


INTROD.]         ANALYTICAL    AND    GEOMETRICAL   MECHANICS.  XXV 11 

abstraction  and  graphical  construction.  The  power  of  abstraction  alone 
suffices,  indeed,  to  comprehend  in  full  generality  metrical  relations  in  ele- 
mentary geometry  and  its  simplest  applications,  but  fails  when  the  relations 
sought  must  be  attained  step  by  step  by  the  application  of  a  number  of 
principles,  or  in  the  auxiliary  figure  by  a  number  of  constructions.  If, 
indeed,  we  take  the  relation  sought  directly  from  the  auxiliary  figure 
itself,  and  even  if  it  were  possible  to  take  out  the  required  distances  with 
absolute  accuracy,  still  this  result  obtained  would  stand  to  the  general  law 
desired  only  in  the  same  relation  that  the  result  of  a  particular  numerical 
computation  does  to  the  more  general  algebraic  formula. 

Investigations  by  the  aid  of  graphical  figures  can,  however,  make  known 
general  relations  of  form  and  position,  and  have  in  this  respect  their  special 
advantage.  So  far  also  as  by  them  metrical  relations  are  sought,  then,  by 
the  exclusion  of  algebraic  formulae,  only  the  process  of  deduction — the 
routine  of  construction — remains  of  general  significance.  Sciences,  then, 
which  proceed  in  this  manner,  furnish  indeed,  with  respect  to  metrical 
relations,  no  general  laws,  but  for  the  deduction  of  these  relations  do  give 
general  methods.  In  this  category  we  may  place  descriptive  geometry  and 
the  more  recent  graphical  statics. 


-it 

ANALYTICAL  AND   GEOMETRICAL   MECHANICS. 

It  is  hardly  necessary  in  these  days  to  call  attention  to  the  advantages 
of  a  geometrical  treatment  of  mechanical  problems.  This,  however,  was 
not  always  the  case,  and  the  most  important  developments  of  geometrical 
mechanics  belong  to  the  present  century.  It  is  to  Poinsot,  Chasles,  Afobius, 
etc.,  that  these  developments  are  due. 

By  the  Calculus  of  Newton  and  Leibnitz  (1646-1714),  and  its  subsequent 
development,  analysis  became  such  a  powerful  instrument  that  the  activity 
of  mathematicians  was  for  a  long  time  solely  directed  towards  analytical 
investigations.  The  power  of  analysis  was  in  mechanics  carried  to  its 
highest  point  by  Lagrange  (1736-1813),  in  his  Mechanique  analytique.  He 
undertook  the  problem  of  reducing  mechanics  to  a  series  of  analytical 
operations :  "  On  ne  trouvera  point  de  figures  dans  cet  outrage.  Les 
methodes  que  fy  expose  ne  demandent  ni  constructions  ni  raisonnement  geo- 
metrique  ou  mecanique,  mais  seulement  des  operations  algebriques  assujeties 
a  une  marche  reguliere  et  uniforme"  (Mechanique  analytique.  Paris,  1788.) 
The  principle  of  virtual  velocities  formed  his  point  of  departure.  A 
number  of  text-books  upon  theoretical  mechanics  still  follow  the  method 
of  Lagrange. 

The  revival  of  pure  geometrical  investigations  by  Monge  (1746-1818), 
the  creator  of  descriptive  geometry,  and  his  followers,  could  not  well  have 
been  without  its  influence  upon  mechanics.  In  the  year  1804  appeared 
the  Elements  de  Statique,  by  Poinsot,  in  which,  in  contrast  to  Lagrange, 


XXviii  ANALYTICAL   AND    GEOMETRICAL   MECHANICS.        [iNTKOD. 

we  find :  "  que  tons  les  theoremes  de  la  Statique  rationeUe  ne  sont  plus  au 
fond  que  des  theoremes  de  Geometric"  This  work  was  the  beginning  of  a 
series  of  treatises  in  which  the  advantages  of  the  synthetic  development 
and  geometrical  treatment  of  mechanics  were  defended  and,  by  most 
important  results,  strikingly  demonstrated. 

At  this  time  the  views  as  to  the  best  method  of  treating  mathematical 
problems  were  sharply  opposed.  Carnot  (1753-1823),  to  whom,  however, 
the  modern  geometry  itself  owes  no  slight  impulse,  gives  the  preference 
to  analysis.  For  synthesis  "  est  restreinte  par  la  nature  de  ces  procedes  ; 
elle  ne  peut  jamais  perdre  de  vue  son  objet,  ilfaut  que  cet  objet  s^offre  tou~ 
jours  d  V esprit,  reel  et  net,  ainsi  que  tons  les  rapprochements  et  combinaisons 
qu'on  en  fait"1  (Oeometrie  de  position.  Paris,  1803.)  That  which  here  Car- 
not considers  as  a  defect  in  the  synthetic  and  geometrical  method,  Poinsot 
claims  as  its  special  advantage  :  "  On  peut  bien  par  ces  calculs  plus  ou  moins 
longs  et  compliques  parvenir  d  determiner  le  lieu  ou  se  trouvera  le  corps  au 
bout  d'un  temps  donne,  mais  ou  le  perd  entierement  de  vue,  tandis  qu'on  vou- 
drait  V observer  et  le  suivre,  pour  ainsi  dire,  des  yeux  dans  tout  le  cours  de  sa 
rotation'1'1  (Theorie  nouv.  d.  I.  rot.  d.  corps}. 

The  example  of  Poinsot  found  numerous  followers.  In  Germany,  Mo- 
bius  followed  with  his  "  Lehrbuch  der  Statik."  Mechanics  as  well  as 
geometry  thus  received  enrichment.  Mobius  gives  the  preference  always 
to  the  synthetic  method,  and  also  endeavors  to  interpret  geometrically, 
analytically  deduced  formulae — "  because  in  investigations  concerning 
bodies  in  space  the  geometrical  method  is  a  treatment  of  the  subject  itself, 
and  is  therefore  the  most  natural,  while  by  the  analytical  method  the  sub- 
ject is  concealed  and  more  or  less  lost  sight  of  under  extraneous  signs  " 
(Lehrb.  d.  StatiTc.  Leipzig,  1837.) 

Even  in  analytical  operations,  geometrical  considerations  came  more  and 
more  in  the  foreground.  On  all  sides  the  development  of  Kinematics,  the 
theory  of  motion  without  reference  to  its  cause,  was  prosecuted.  But. 
neglecting  the  cause  of  motion,  there  remains  only  its  path ;  that  is,  geo- 
metry proper  (Kinematical  geometry,  or  the  geometry  of  motion}.  The  in- 
vestigations of  Chasles,  Mobius,  Rodrigues,  Jouquiere,  and  others,  may  yet 
be  still  further  pursued ;  and  when  by  the  aid  of  geometry  a  certain  com- 
pleteness has  been  given  to  the  theory  of  the  motion  of  invariable  systems, 
the  geometrical  theory  of  regular  variable  systems  (to  which  the  flexible 
and  elastic  belong)  will  be  possible.  For  the  discussion  of  such  branches 
of  mathematics,  the  synthetic  geometry  is  necessary  ;  for  their  foundation 
lies  in  a  theory  of  the  relationship  of  systems. 

The  advantage  of  the  synthetic  method  in  mechanics  is  denied  by  no 
one.  Wherever  it  is  possible,  we  obtain  more  comprehensive  conclusions 
as  to  the  nature  of  the  phenomena,  while  all  the  properties  of  the  same  fol- 
low directly  from  the  simple  and  known  truths  premised.  In  analytical 
investigations  it  is  necessary,  even  when  definite  equations  are  obtained,  to 
deduce  the  actual  laws  singly  and  in  a  supplementary  manner,  although 
they  are  indeed  all  contained  in  the  equations  themselves. 

It  is  not,  however,  always  possible  to  preserve  the  synthetic  process 
throughout.  From  the  first  truth  the  ways  diverge  in  all  directions,  and 


TNTROD.]  GEOMETRICAL    STATICS.  Xxlx 

a  special  ingenuity  is  often  needed  to  reach  the  goal.  Just  here  analysis 
comes  to  our  aid  with  its  rich  treasures  of  developed  methods,  and  here  it 
is  most  certainly  not  for  geometry  to  "  undervalue  the  advantage  afforded 
by  a  well-established  routine,  that  in  a  certain  degree  may  even  outrun  the 
thought  itself  "  (F.  Klclu  :  Vrrglaichemle  JRetrachtungen  aber  neuere  geome- 
trische  Forschung^en.  _  Erlanggn-,  1  ft72T  p.  41).  Algebraic  operations  are 
thus,  however,  not  the  chief  thing,  but  only  the  instrument — a  most  excel- 
lent instrument  indeed,  which  can  be  almost  universally  applied,  and 
which,  by  reason  of  its  connection  with  an  extensive  and  independent 
mechanism,  often  needs  only  to  be  set  in  action  in  order  to  work  of  itself. 

Geometrical  mechanics,  moreover,  can  never  entirely  free  itself  from 
analytical  formulae  and  operations.  For  though  it  may  be  both  interesting 
and  useful  to  follow,  with  Poinsot,  the  body  during  its  entire  rotation,  yet 
practically  this  is  of  minor  interest,  and  the  chief  problem  remains  still, 
" d  determiner  le  lieu  ou  se  trouvera  le  corps  au  T)out  cTun  temps  donne" 

In  the  present  day  all  those  familiar  with  both  methods  of  treatment 
hold  fast  the  good  in  each ;  they  supplement  each  other.  Often  in  the 
course  of  the  same  investigation  we  must  interrupt  the  general  analytical 
process  with  synthetic  deductions,  and  inversely.  Thus  we  may  well  close 
these  considerations  with  the  sentence  with  which  Schell  begins  his  "  Theo- 
rie  der  Bewegung  und  der  Krdfte  " — both  methods,  the  analytic  and  the 
synthetic,  can  only,  when  united,  give  to  mechanics  that  sharpness  and 
clearness  which  at  the  present  day  ought  to  characterize  all  the  mathemati- 
cal sciences. 


III. 

GEOMETRICAL    STATICS. 

Statics  is  a  special  case  of  dynamics,  though  earlier  treated  as  indepen- 
dent of  the  latter.  The  principle  of  d'Alembert  furnishes  the  means  of 
passing  from  one  .to  the  other.  In  technical  mechanics  the  distinction  is 
still  preserved,  and  indeed,  in  view  of  the  distinct  branches  in  which  the 
applications  on  either  side,  are  found,  not  without  propriety. 

After  the  mechanics  of  the  ancients,  as  comprised  in  the  mathematical 
collections  of  Pappus;  the  first  great  step  towards  our  present  geometrical 
statics  was  made  by  Simon  Stemnus  (1548-1603),  when  he  represented  the 
intensity  and  direction  of  forces  by  straight  lines.  Stevinus  himself  gave 
a  proof  of  the  importance  of  his  method,  in  the  principle,  deduced  from  it, 
that  three  forces  acting  upon  a  point  are  in  equilibrium  when  they  are  pro- 
portional and  parallel  to  the  three  sides  of  a  right-angled  triangle. 

A  main  discovery  was  the  parallelogram  of  forces  by  Newton  (1642- 
1727).  The  composition  of  two  velocities  in  special  cases  was  long  famil- 
iar. Galileo  made  use  of  it  for  two  velocities  at  right  angles,  and  exam- 
ples also  occur  in  Descartes,  Roberval,  Mersenne,  and  Wallis,  but  the  funda- 
mental principle  was  first  established  when  Newton  replaced  the  theories 


XXX  GEOMETRICAL    STATICS.  [iNTEOD. 

of  special  by  that  of  universal  causation  (Philosophic^  naturalis  principia, 
mathematica.     London,  1687). 

Varignon  in  his  "Projet  d'unenouvelle  mecanique,"  in  the  same  year  (1687), 
and  independently  of  Newton,  applied  for  the  first  time  the  general  princi- 
ple of  the  composition  of  motions.  From  this  he  passes,  in  the  Nouvelle 
mecanique  ou  Statique,  dont  le  projet  fat  donne  en  1687  (published  after 
his  death,  Paris,  1725),  by  means  of  the  axiom  that  "  les  effets  sont  toujours 
proportionnels  d  leurs  causes  ou,  forces  productrices  "  to  the  composition  of 
forces  also. 

The  Statique  of  Varignon  is  purely  geometrical.  He  postulates  nothing 
beyond  books  1-6  and  11  of  Euclid,  and  even  explains  the  significance  of 
+  and  —  signs.  In  this  work,  the  first  founded  upon  the  parallelogram  of 
motion  and  of  forces,  we  find  also  the  force  and  equilibrium  polygons 
(Funiculaire,  Section  II.),  to  the  application  and  development  of  which 
almost  the  whole  of  Graphical  Statics  is  to  be  attributed.  Varignon  recog- 
nized the  value  of  the  equilibrium  polygon,  and  gave  it  as  the  seventh  of 
the  simple  machines. 

After  the  great  Interim  of  Geometry,  Monge  wrote  a  Traite  elementaire 
de  Statique  (Paris,  1786).  The  work  claims  to  contain  for  the  first  time 
everything  in  statics  which  can  be  synthetically  deduced.  In  a  later  edi- 
tion we  learn  that  synthetical  statics  must  be  taken  up  as  preliminary  to 
analytical,  just  as  elementary  geometry  before  analytical  geometry.  Thus 
the  work  of  Monge  contains  the  necessary  preparation  for Poisson1  s  "Traite 
de  mecanique"  (Paris,  1811). 

The  greatest  influence  upon  the  development  of  geometrical  statics  was 
exercised  by  Poinsot.  By  the  introduction  of  force  pairs,  he  solved  in  the 
most  elegant  manner  the  fundamental  problem  of  any  number  of  forces 
acting  upon  a  body  (Elements  de  Statique,  Paris,  1804,  and  Memoire  sur 
la  composition  des  moments  et  des  aires  dans  la  mecanique). 

Chasles  completed  the  solution  by  the  proof  that  the  contents  of  the 
tetrahedron,  which  is  determined  by  the  resultant  forces,  is  constant,  how- 
ever the  forces  may  be  composed. 

In  the  hands  of  Mobius,  geometry  and  geometrical  statics  were  most  com- 
pletely developed. 

Of  the  greatest  importance,  for  later  applications,  was  the  introduction 
of  the  rule  of  signs. 

The  germ  of  this  had  existed  already  in  the  preceding  century.*  Mobius 
recognized  its  significance,  extended  it  to  the  expression  of  the  contents  of 
triangles,  polygons,  and  three-sided  pyramids,  and  applied  it  systemati- 
cally (Barycentrischer  Calcul.  Leipzig,  1827). 

A  new  impulse,  extended  field  of  action,  and  numerous  additions  were 
given  to  geometrical  statics  by  the  Graphical  Statics  of  Culmann. 

*  Mobius  alludes  to  this,  and  we  find,  for  example,  in  Kastner  ( Geometrische 
Abhandlungen,  I.  Saml.,  1790,  p.  464),  the  equation  A  B  +  B  A  =  o. 


INTROD.]  THE    GRAPHICAL   CALCULUS.  XXxi 


IV. 

THE   GRAPHICAL   CALCULUS. 

The  most  extended  applications  of  statics  are  in  the  Held  of  engineering. 
Here,  not  only  general  properties  of  form  and  position  are  required,  but  in 
a  large  number  of  cases  numerical  relations  are  also  necessary.  General 
results  of  the  latter  character  can,  as  we  have  seen,  only  be  embraced  by 
algebraic  formulae  (I).  The  pure  graphical  theory  of  construction  is  there- 
fore in  this  respect  lacking  in  completeness,  as  it  is  unable  to  furnish  gen- 
eral metrical  relations. 

The  practical  engineer  has  almost  always,  however,  to  do  with  special 
problems;  dimensions  and  acting  forces  are  numerically  given.  Geometry 
in  such  cases  could  give  no  general  relations,  because  the  results  desired 
are  the  consequences  of  the  special  proportions  of  the  figure.  In  any  de- 
terminate case,  however,  we  may  obtain  a  result  holding  good  for  that  case, 
and  it  only  remains  to  show  how  generally  to  obtain  such  a  result.  The 
graphical  calculus  treats  of  such  methods,  and  so,  although  not  exclusively, 
does  graphical  statics.  As  soon  now  as  practical  use  is  made  of  the  actual 
proportions  of  the  figure,  everything  depends  upon  the  exactness  of  the 
drawing.  One  condition  for  the  application  of  the  graphical  method  is, 
therefore,  skill  in  geometrical  drawing — a  requisition,  indeed,  which  the 
practical  engineer  can  most  readily  meet. 

The  idea  at  bottom  of  the  graphical  calculus  is  simple.  The  modifica- 
tions of  numbers  in  numerical  calculations  correspond  always  to  similar 
modifications  of  the  quantities  represented  by  these  numbers.  The  measure 
of  a  quantity  can  be  as  well  given  by  a  line  as  a  number,  by  putting  in 
place  of  the  numerical  the  linear  unit.  In  order  for  a  graphical  calculus, 
then,  the  modifications  of  lines  answering  to  corresponding  numerical 
operations  are  necessary,  and  these  are  furnished  by  geometry.  They  con- 
sist of  graphical  constructions,  and  rest  upon  the  known  properties  of 
geometrical  figures.  The  scale  furnishes  the  means  of  converting  directly 
any  numerical  quantity  into  its  corresponding  linear  representation,  and 
inversely  any  graphically  obtained  result  can  be  at  once  transformed  into 
numbers. 

The  graphical  determination  of  desired  or  computable  numbers  is  natu- 
rally nothing  new.  From  the  "  Traite  de  Q-nomonique "  of  de  la  Hire 
(1682)  to  the  "  Geometric  descriptive'1'1  of  Monge  (1788),  many  examples 
are  to  be  found.  The  graphical  calculus,  however,  goes  further  than  this. 
It  aims  to  found  a  method,  a  routine,  which  shall  not  only  apply  to  bodies 
in  space,  but  which  shall  also,  like  the  arithmetical  or  algebraic  calculus, 
be  independent  of  concrete  relations  and  of  general  application.  It  seeks 
further  to  obtain  its  results  (products  and  powers)  in  the  shape  of  lines 
convertible  by  scale  into  numbers.  (Hence  the  important  part  which  area 


XXXM  THE    GRAPHICAL    CALCULUS.  [iNTROIX 

transformation  plays  in  the  graphical  calculus.)  Such  was  the  problem 
which  Cousinery  proposed,  and  whose  solution  he  attempted  in  his  "  Calcul 
par  le  trait"  (Ses  Elements  et  ses  applications.  Paris,  1839). 

Cousinery  applied  the  graphical  calculus  to  powers,  roots,  proportion 
and  progression;  to  the  measure  of  lines,  surfaces,  cubes,  graphic  inter- 
polation, and  the  strength  of  retaining  walls.  The  presentation  is  nat- 
urally by  no  means  complete,  and  labors  also  under  a  prolixity  and 
minuteness  of  detail  to  which  the  results  obtained  are  by  no  means  com- 
mensurate. It  sounds  somewhat  comic  when  Cousinery,  in  his  "  Calcul  par 
le  trait,"  claims  the  then  already-existing  graphical  solutions  of  Poncelet 
(" Memoire  sur  la  stdbilite  des  revetements,  in  Memorial  de  Toff  du  genie") 
as  an  elegant  example  of  the  application  of  his  graphical  calculus. 

While  Cousinery  thus  sought  to  apply  geometry  in  a  direction  where 
until  then  analysis  had  held  sway,  he  acted  in  entire  accordance  with  the 
spirit  of  his  age,  though  without  making  use  of  those  means  for  aid  which 
lay  at  his  disposal.  "  Without  effect  upon  him,"  says  Culmann,  "  were 
the  researches  of  Steiner,  already  published  in  1832,  as  well  as  those  of 
his  predecessor;  and  instead  of  simply  premising  the  elementary  prin- 
ciples of  the  modern  geometry,  he  laboriously  sought  to  deduce  them  in- 
dependently by  the  aid  of  perspective."  The  works,  at  least,  of  the  French 
predecessors  of  Steiner  were,  at  any  rate,  well  known  to  Cousinery.  In  his 
preface  we  read:  "  Peut-gtre  mihne  nos  efforts  eussent-ils  6t6  complete- 
ment  infructueux,  sans  les  ressources  que  nous  ont  procurers  et  les  annales 
de  M.  Gergonne  et  les  travaux  de  M.  Brianchon,  et  ceux  plus  re"cents  de 
M.  Poncelet.  Nous  avons  envers  M.  Chasles  une  obligation  encore  plus 
droite,  car  outre  les  pr6cieux  documents  que  reufernie  son  '  Histoire  des 
methodes  en  geometric?  nous  avons  a  lui  faire  agre"er  un  tgmoignage  par- 
ticulier  de  reconnaissance  pour  la  maniere  dont  il  a  bien  voulu  mentionner 
nos  premiers  essais  sur  le  systeme  de  projection  polaire." 

Why  Cousinery  made  use  of  perspective  and  not  of  the  modern  geome- 
try, is  easily  understood.  The  development  of  geometry  at  that  time,  as  to- 
day, proceeded  in  various  almost  independent  directions,  and  Cousinery 
himself  had  the  pleasure  of  seeing  his  "  Geometric  perspective "  (Paris, 
1828)  designated  by  the  reporters  for  the  Academy,  Fremel  and  Matthien, 
as  new  and  ingenious,  as  well  as  favorably  noticed  by  Chaxles.*  He 
sought,  therefore,  naturally  to  develop  and  render  fruitful  his  own  method, 
so  much  the  more  as  the  true  significance  and  value  of  the  various  growing 
branches  of  geometry  could  not  then,  as  now,  be  correctly  estimated.  Ac- 
cordingly, the  Inge"nieur-en-chef.  B.  E.  Cousinery,  wrote  avowedly  for  his 
colleagues,  and  did  not  feel  justified  in  directly  premising  a  knowledge  of 
the  newest  investigations,  more  especially  of  his  own. 
i  We  have  noticed  the  above  somewhat  in  detail,  because  it  bears  directly 

*  Its  newness,  at  least,  is  not  without  doubt.  According1  to  Fiedler,  the 
principles  are  completely  given  in  Lambert's  celebrated  work,  "  Die  freie 
Perspective"  (Zurich,  1759).  Poncelet  also  takes  issue  with  the  estimation  of 
the  "  Geometrie  perspective"  by  Chasles  ("  Traite  des  propr.  proj.^  II. ,  ed. 
1865,  p.  412). 


INTKOD.]  THE    GRAPHICAL    CALCULUS.  XXXI 1 1 

upon  a  point  of  our  discussion ;  for  the  introduction  of  the  modern 
geometry  in  the  graphical  method  by  Culmann,  is  still,  thirty  years  after 
Cousinery,  a  chief  hindrance  to  its  rapid  spread.* 

After  Cousinery,  no  one  occupied  himself  with  the  graphical  calculus 
till  Culmann  gave  it  a  place  in  his  GrapJiische  Statih  The  presenta- 
tion is  here  far  better,  and  especially  shorter.  The  rule  of  signs,  which 
was  unknown  to  Cousinery,  is  at  once  brought  out.  Instead  of  such 
long  and  tedious  applications  as  the  graphical  interpolation,  a  few 
examples  from  engineering  practice  are  given,  among  which  we  may 
especially  notice  earth-work  calculations.  In  the  extensive  earth  works  of 
roads,  canals,  and  railways,  the  method  shows  not  only  most  plainly  the 
extent  and  best  arrangement  of  transport,  but  also  allows,  with  the  aid  of 
the  planimetre,  the  cost  of  transport  to  be  determined. 

As  to  the  rest,  it  would  appear  as  if  the  graphical  calculus  should  play 
an  important  part  in  engineering  practice.  This  circumstance,  as  well  as 
the  interesting  problems  which  present  themselves  in  connection,  has 
gained  for  the  Arithmograpliy  many  friends.  Several  publications 
have  since  sought  to  win  for  it  a  wider  recognition  without  furnishing 
anything  essentially  new.  [IT.  Eggers :  "  Grundziige  einer  graphischen 
Arithmetic,"  Schaffhausen,  1865.  J.  ScUesinger :  "  Ueber  Potenzcurven," 
Zeitschr.  d.  osterr.  Arch.  u.  Ing.  Ver.,  1866.  E.  Jdger  :  "  Das  graphischen 
Rechnen,"  Speier,  1867.  K.  von  Ott :  "  Grundziige  des  graphischen  Rech- 
nens  und  der  graphischen  Statik,"  Prag,  1871.] 

Recently  the  method  of  the  graphical  calculus  has  been  applied  to  Dif- 
ferentiation and  Integration.  A  treatise  by  Solin  show§  the  first  exact,  so 
far  as  possible  in  a  construction,  the  last  approximate  only  ("  Ueber  graph. 
Integr.  ein  Beitrag  z.  Arithmographie,  Abhand.  d.  konigl.  bohm.  Gesellsch. 
d.  Wissenbach."  VI.  Folge,  5  Bd.  Separate  reprint  by  Rivnac,  Prag, 
1871).  It  is  to  be  remarked,  also  that  examples  of  double  integration  and 
differentiation  were  given  by  Mohr  in  1868.  The  graphical  construction 
of  the  elastic  line,  and  the  determination  of  the  moments  at  the  supports 
of  a  continuous  girder,  are  essentially  examples  in  point  (Mohr:  "Bei- 
trag zur  Theorie  der  Holz  und  Eisenconstructionen,1'  Zeitschr.  d.  Hannov. 
Ing.  und  Arch.  Ver.,  1869  ;  or  W.  Hitter  :  "  Die  elastische  Linie,"  Zurich, 
1871.) 

As  to  the  importance  of  the  graphical  calculus  as  an  independent  study 
or  discipline,  it  is,  as  we  believe,  often  exaggerated.  The  theoretical  value  is 
but  little,  and  for  graphical  constructions,  as  given  by  the  graphical  calculus, 
offer  in  no  respect  anything  new.  That  which  pertains  to  practical  applica- 
tions may  be  easily  based  directly  upon  geometry,  and  is  nowhere  found 
as  a  consequence  of  the  method  itself.  If  it  is  considered  advisable  to  call 
special  attention  to  a  few  general  points  before  making  such  applications, 
all  that  can  be  desired  can  be  easily  presented  in  ten  or  a  dozen  pages 
octavo. 

*  See  Preface  ;  also  Chaps.  VII.  and  VIII.  of  this  Introduction. 
c 


GRAPHICAL   REPRESENTATION.  [iNTROD. 


V. 

GRAPHICAL   REPRESENTATION. 

Graphical  representation,  in  the  widest  sense  of  the  word,  includes  every 
visible  result  of  writing  or  drawing.  The  written  sentence  is  the  graphi- 
cal representation  of  a  thought — the  drawn  line  the  graphical  indication  of 
an  idea.  In  such  generality  we  naturally  do  not  here  regard  graphical 
representation.  In  a  narrower  sense  we  understand  the  graphical  represen- 
tation of  the  diversity  or  dependence  of  numerical  quantities.  In  this 
sense  we  cannot  speak  of  the  graphical  represention  of  pure  geometry. 
This  last  was  introduced  into  analysis  by  Vieta  (1540-1603).  Here 'the 
figure  merely  aids  the  conception,  while  the  equation  embraces  the  charac- 
teristics of  the  phenomena  (I.),  and  ensures  the  independent  character  of 
the  drawn  lines.  Thus  the  clearness  of  geometry  is  combined  with  the 
fruitf ulness  of  analysis. 

If  the  graphical  representation  is  constructed  frcm  a  number  of  suitably 
chosen  and  calculated  values,  the  intermediate  values  can  be  directly  meas- 
ured and.  by  means  of  the  scale,  reconverted  into  numbers.  The  graphical 
representation,  then,  replaces  numerical  tables.  Illustrative  examples  often 
occur  in  practice.  We  instance,  for  example,  the  graphical  representation 
of  maximum  moments  and  shearing  forces  in  the  continuous  girder.  If 
the  several  values  are  calculated  from  a  formula,  their  graphical  union  gives 
a  simultaneous  view — a  picture — of  the  law  which  the  formula  represents. 
If  these  values  are  merely  known — observed,  for  example  their  graphical 
combination  may  enable  us  to  deduce  the  law  which  connects  them.  Thus 
the  graphical  representation  is  of  assistance  in  the  deduction  of  empirical 
formulae,  and  indirectly  in  the  discovery  of  exact  relations.  Illustrations 
of  such  application  occur  frequently  in  applied  mathematics,  especially  in 
astronomy  and  meteorology. 

In  this  connection  we  may  also  remark  that  graphical  representation 
plays  also  an  important  part  in  statistics.  By  its  aid  a  comprehensive  view 
is  obtained  of  a  series  of  separate  results.  Or  it  may  be  applied  to  still 
higher  problems — for  example,  from  comparison  of  simultaneous  but  differ- 
ent series  of  observations  to  determine  an  inner  connection. 

In  engineering  practice,  graphical  representations  have  in  recent  times 
notably  multiplied.  All  graphical  constructions,  so  far  as  they  do  not  de- 
pend upon  analytical  formulae,  and  therefore  are  not  directly  given  by 
geometrical  laws,  are  nothing  more  than  consequences  of  graphical  repre- 
sentation. 


INTROD.]  GRAPHICAL    STATICS.  XXXV 


VI. 

GRAPHICAL    STATICS. 

The  few  text-books  upon  graphical  statics  and  the  more  numerous  works 
upon  its  applications,  afford  us  no  definition,  and  can  afford  none,  because 
neither  the  method  nor  scope  of  this  new  study  are  anywhere  sufficiently 
indicated. 

If,  following  Culmann,  we  speak  of  it  in  contradistinction  to  the  appli- 
cations of  a  pure  graphical  statics,  we  may  define  it  somewhat  as  follows : 
Graphical  Statics  comprises  the  theory  of  those  geometrical  constructions  which 
occur  in  the  graphical  solution  of  statical  engineering  problems ;  it  treats 
further  of  the  general  relations  deducible  from  such  constructions.  This 
limitation,  so  far  as  it  does  not  follow  from  the  preceding,  we  shall  seek 
in  the  course  of  these  remarks  still  further  to  establish. 

Graphical  representations  of  analytically  obtained  results  have,  as  has 
been  already  noticed,  long  been  used  in  engineering  practice.  They  served 
also  the  purposes  noticed  in  the  preceding  chapter.  Often  also  certain 
values,  whose  analytical  determination  is  somewhat  complicated,  have 
been  sought  by  graphical  constructions.  Examples  of  this  may  be  found 
in  many  text-books  upon  the  theory  of  structures,  and  we  notice  only,  as 
one  of  the  most  notable  of  recent  date,  the  construction  of  lever  arms  and 
limits  of  loading  in  A.  Ritter's  "Theorie  und  Berechnung  eiserner  Dach 
und  Briickenconstructionen "  (Hannover,  1862).  Poncdet  applied  analy- 
sis in  general  to  practical  investigations,  but  sought  in  several  complicated 
cases  to  elucidate  the  deductions  of  formulae  by  geometrical  constructions, 
and  to  deduce  graphical  solutions  from  analytical  relations.  This  pro- 
cedure found  considerable  acceptance,  and  the  investigations  of  Poncelet 
were  afterwards  resumed  upon  more  general  assumptions  by  Saint  Guil- 
liem  (Memoir  e  sur  la  poussee  des  terres  avec  on,  sans  surcharge,  aim.  des 
ponts  et  chauss.,  1858,  sem.  1,  p.  319). 

The  first,  however,  to  give  pure  geometrical  determinations  of  stability 
in  structures  was  Cousinery.  He  gave  a  number  of  examples  as  applica- 
tions of  his  graphical  calculus,  but  his  ideas  appear  to  have  found  in 
France  little  acceptance.  On  the  other  hand,  the  graphical  construction  of 
the  curve  of  pressure  in  the  arch  by  Mery  (Memoir e  sur  V equilibre  des  voutes 
en  bwceau — ann.  d.  ponts  et  chauss.,  1840,  sem.  1,  p.  50)  was  extensively 
used,  and  has  since  been  extended  by  Durand-Claye  to  iron  arches  also 
(Ann.  d.  ponts  et  chauss.,  1867,  sem.  1,  p.  63,  and  1868,  sem.  1,  p.  109). 
Special  prominence  was  given  to  graphical  investigations  of  stability  by 
Culrnann's  "  Graphische  Statik "  (first  part,  Zurich,  1864,  entire  work, 
1866  ;  second  edition,  1st  part,  1875.) 

This  work  of  Culmann  must  be  considered  as  original  in  all  those  parts 
relating  to  structures.  Poncelet  and  Cousinery,  beyond  the  general  idea, 
furnished  only  unessential  contributions.  Culmann  recognized  the  fruit- 


XXXvi  GRAPHICAL    STATICS.  [iNTROD. 

fulness  of  the  relations  between  the  force  and  equilibrium  polygon,  upon 
which  most  of  the  practical  solutions  depend.  He  developed  these  rela- 
tions, applied  them  in  the  theory  of  moments  by  the  introduction  of  the 
closing  line  (Schluss  Linie),  and,  accepting  the  rule  of  signs,  obtained  gen- 
eral points  of  view  for  the  discussion  of  the  most  diverse  figures  which 
could  arise  in  the  same  problem.  In  this  and  in  many  other  respects  even 
geometrical  statics  can  profit  from  Culmann' s  work,  as,  for  instance,  in  the 
investigation  of  the  projective  relations  between  the  force  and  equilibrium 
polygon. 

The  fundamental  importance  of  the  force  and  equilibrium  polygon  was 
also  recognized  by  those  who,  after  Culmann,  occupied  themselves  with 
the  graphical  method.  Here  we  may  notice  two  works  of  special  influence 
upon  the  development  of  the  graphical  statics— those  of  MoTir  and  Cre- 
mona. The  idea  of  Molir,  that  the  elastic  line  is  an  equilibrium  polygon  or 
curve  ("  Beitrag  zur  Theorie  der  Holz  und  Eisenconstructionen."  Zeitschr. 
d.  Hannov.  Ing.  und  Arch.  Ver.,  1868)  is  of  special  significance  forgraphi. 
cal  statics. 

That  from  it  MoJir  obtained  the  graphical  determination  of  the  moments 
at  the  supports  of  a  continuous  girder,  is  an  example  both  useful  as  well 
as  interesting.  Already  it  has  been  endeavored  to  utilize  the  same  idea  in 
other  cases  (Frankel :  "  zur  Theorie  der  Elastischen  Bogentrager,"  Zeitschr. 
d.  Hannov.  Ing.  u.  Arch.  Ver.,  1869,  p.  115),  and  by  it  an  impulse  has  been 
given  to  similar  investigations. 

Cremona,  has  kept  more  especially  in  view  the  geometrical  side  of  graphi  • 
cal  statics.  Starting  from  the  theory  of  reciprocal  polyhedrons,  he  gave 
the  reciprocal  relations  between  the  force  and  equilibrium  polygon  with  a 
generality  and  elegance  to  be  expected  from  this  distinguished  Italian 
mathematician  (Le  figure  reciproche  nelle  statica  graficu,.  Milan,  Langer, 
1872).  By  this  investigation  the  theoretical  development  of  the  graphical 
statics  is  essentially  anticipated. 

It  was  under  the  most  unfavorable  circumstances  that  Culmann  intro- 
duced his  graphical  statics  in  the  engineering  department  of  the  Ziirich 
Polytechnic  in  the  year  1860.  It  was  finally,  indeed,  admitted  as  a  regular 
study,  but  not  the  geometry  of  position  which  he  premised.  It  was  not 
till  1864  that  this  last  was  given  in  a  series  of  lectures  by  Reye,  and  then 
the  time  at  disposition  for  both  courses  was  insufficient.  Meanwhile  the 
method  spread,  crept  into  the  construction  department  of  the  engineering 
school,  and  wherever  it  came,  even  in  the  other  departments  of  the  Poly- 
technic, gained  friends.  Finally,  at  the  present  time,  it  is  to  be  found,  to- 
gether with  the  modern  geometry  of  position,  upon  which  it  was  based,  in 
every  Polytechnic  throughout  Germany. 

According  to  the  above  given  definition  of  graphical  statics,  the  methods 
of  the  graphical  calculus,  as  far  as  applied  in  statical  investigations,  may 
also  be  regarded  as  belonging  to  graphical  statics,  and  justly  so ;  for 
these  methods  follow  directly  from  geometrical  principles,  and  can  be  ap- 
plied by  any  one  acquainted  with  geometry,  without  being  collected  under 
the  special  name  of  the  "  graphical  calculus."  Thus,  for  instance,  Bausch- 
inger,  in  his  "  Elemente  der  graphischen  Statik"  (Miinchen,  1871),  disre- 


INTROD.]       METHODS  AND    LIMITS    OF    GRAPHICAL    STATICS.          XXXvil 

gards  entirely  the  graphical  calculus,  and  also  cuts  loose  from  the  modern 
geometry ;  he  develops  the  elementary  principles  of  the  subject  in  a  logi- 
cal and  easily  comprehended,  if  not  purely  geometrical  manner,  and  thus 
brings  the  subject  within  the  reach  of  those  persons  for  whom  it  seems  so 
especially  designed.  The  work  is  remarkable  for  clear  presentation,  but 
expressly  avoids  all  special  investigations  and  practical  applications,  for 
which  it  is  merely  intended  to  prepare  the  way.  In  the  present  work,  also, 
a  similar  plan  is  pursued,  but  all  such  applications  as  are  of  most  value  to 
the  engineer  or  mechanic  find  likewise  a  place.  Thus,  combining  the 
method  of  presentation  of  Bauschinger  and  the  practical  applications  of 
(Julmann,  it  has  been  endeavored  to  make  it  a  practical  manual,  as  well  as 
a  text-book  of  elementary  principles — to  serve  the  wants  of  the  practical 
engineer,  and  also  meet  the  requirements  of  the  engineering  student.  How 
far  this  twofold  design  has  been  realized,  the  judgment  of  the  reader 
must  decide. 


VII. 

THE   METHODS   AND   LIMITS    OF    THE    GRAPHICAL    STATICS. 

The  most  perfect  method  of  the  graphical  statics  is  the  synthetic  or  geo- 
metric; since  in  geometrical  statics  the  solution  must  always,  when  possi- 
ble, rest  upon  pure  mechanical  or  geometrical  reasoning.  Culmann  pre- 
sents his  graphical  statics  to  practitioners  "as  an  attempt  to  solve  by  the 
aid  of  the  modern  geometry  such  problems  pertaining  to  engineering  prac- 
tice as  are  susceptible  of  geometrical  treatment." 

The  graphical  statics,  however,  is  not  in  and  of  itself  the  product  of 
endeavors  to  make  the  modern  geometry  of  service  in  applied  mechanics  ; 
graphical  solutions  merely  were  required.  How  to  obtain  these,  was 
another  question.  Thus  it  is  that  Poncelet's  solutions  consist  almost  en- 
tirely of  graphical  representations  of  analytical  relations ;  that  Cousinery 
avoided  all  use  of  formulae ;  that  Culmann  made  use  of  the  new  geometry 
wherever  it  was  possible ;  that  Bauschinger  and  others  make  use  only  of 
the  ancient  geometry ;  and  that  the  latest  graphical  solutions — in  a  certain 
degree,  those  of  Mohr  also — entirely  in  the  spirit  of  Fencelet's,  rest  again 
upon  analysis.  The  pure  geometric  solution  is,  indeed,  desirable,  but  is  not 
always  attainable. 

If  now  we  review  all  the  cases  in  which  direct  and  exclusively  geomet- 
rical solutions  are  not  possible,  we  see  at  once  that  this  occurs  when  it  is 
required  to  make  use  of  the  physical  properties  of  bodies,  as  elasticity,  co- 
hesion, etc.  Why?  The  actual  condition  of  a  body  after  equilibrium  18 
attained,  is  a  consequence  of  the  motion  of  a  variable  system  of  points. 
The  theory  of  the  motion  of  variable  systems  has,  however,  by  no  means,  as 
yet,  been  brought  to  practical  efficiency  (II.).  We  are  therefore  obliged  to 
start  from  an  hypothetical  condition  or  state  of  the  body  (in  the  theory  of 
flexure,  for  instance,  we  rest  upon  the  assumption  that  all  plane  cross-sec- 
tions made  before  the  action  of  the  outer  forces  remain  plane  after  their 


XXXVlii          METHODS  AND  LIMITS  OF  GRAPHICAL  STATICS.         [iNTEOD. 

action).  To  deduce  now  from  this  general  condition  the  special  relations 
necessary  for  solution,  demands  an  essentially  analytical  process  (T) 
Hence  the  dependence  of  the  graphical  solutions  in  such  cases  upon  ana- 
lytical relations — relations  which,  when  the  body  is  assumed  to  be  rigid, 
as  in  the  arch,  in  frame  work,  or  the  simple  girder,  no  longer  exist. 

The  sphere  of  action  of  an  independent  graphical  statics  is,  then,  con- 
fined to  those  problems  which,  under  the  assumption  of  inflexibility,  are 
determined  by  a  sufficient  number  of  conditions.  Beyond  this  point  we 
have  chiefly  graphical  interpretations  only. 

It  has  been  already  noticed  that  graphical  statics,  without  the  application 
of  algebraic  operations,  can  furnish  no  general  laws  (IV.).  From  relatively 
simple  figures,  indeed,  here  and  there,  general  formula  of  metrical  relations 
have  been  derived,  as  is,  in  fact,  not  theoretically  impossible  (I.),  but  such 
formulae  were  always  previously  known.  Such  a  result  holds,  in  general, 
immediately  good  only  for  that  form  of  figure  which  has  been  discussed, 
Or,  according  to  the  terminology  of  Carnot,  only  for  the  existing  "  primi- 
tive figure,"  and  must  be  proved  or  transformed  for  all  "  correlative 
figures  "  which  can  occur  in  accordance  with  the  conditions  of  the  prob- 
lem. When  the  graphical  investigation  is  guided  by  analytical  opera- 
tions, it  is  these  last  which  render  possible  the  deduction  of  general  metri- 
cal relations. 

Thus,  in  the  theory  of  structures,  there  remains  subject  to  pure  graphical 
treatment  only  the  general  relations  of  form  and  position.  Here  we  have 
the  elegant  deductions  upon  unfavorable  loading,  and  here  the  graphical 
method  often  attains  its  end  in  a  more  elegant  manner  than  the  analytical. 
A  complete  exploration  and  development  of  such  form  and  place  relations, 
without  a  geometry  of  position,  would  evidently  be  impossible  (IX.).  The 
scientific  future  of  the  graphical  statics,  therefore,  rests  essentially  upon 
the  influence  of  the  modern  geometry.  To  endeavor  to  separate  the  higher 
geometry  from  the  graphical  method  would  be  as  unwise  and  fruitless  as 
the  attempt  to  exclude  the  higher  analysis  from  analytical  investigations. 
As,  however,  for  certain  purposes  an  elementary  presentation  of  analytical 
theories  relating  to  engineering  practice  will  ever  be  acceptable,  so  also  an 
elementary  development  of  graphical  methods  is  not  without  justification, 
the  more  so  as  long  as  the  modern  geometry  itself  is  not  sufficiently  well 
known. 

Culmann  says  of  the  graphical  statics :  "  It  includes,  thus  far,  only  the 
general  part  which  we  need  in  the  investigation  of  problems  in  construc- 
tion, but  it  must  and  will  extend,  as  graphical  methods  find  ever  wider 
acceptance.  Then,  however,  it  will  escape  the  hands  of  the  practitioner, 
and  must  be  built  up  by  the  geometer  and  mechanic  to  a  symmetrical 
whole,  which  shall  bear  the  same  relation  to  the  new  geometry  that  analyti- 
cal mechanics  does  to  the  higher  analysis."  Such  an  estimation  does  not 
appear  to  be  entirely  correct.  It  is  geometrical  statics  (or  mechanics)  for 
which  the  above  relation  may  subsist,  and  to  this,  indeed,  Culmann's  valu- 
able work  has  itself  greatly  contributed.  It  was,  moreover,  developed 
quite  independently  of  and  much  earlier  than  graphical  statics  (III.).  In 
this  respect,  therefore,  the  spread  of  graphical  methods  is  of  less  inipor- 


INTROD.]  THE   MODERN   GEOMETRY. 

tance  than  that  of  geometrical  views  and  knowledge ;  for  when  practical 
calculations  are  disregarded,  and  the  deduction  of  general  truths  alone 
occupies  us,  then,  first  of  all,  we  must  exclude  from  the  drawn  figure  all 
special  relations — that  is,  strike  out  of  graphical  statics  the  essentially 
graphical  part.  A  truth  comprehended  only  in  the  abstract  holds  good 
for  all  figures  which  can  be  drawn  in  accordance  with  the  given  condi- 
tions. 

We  place,  then,  in  one  line  geometry  and  geometrical  statics  (mechanics). 
From  geometry  we  obtain  a  method  of  construction,  or  descriptive  geome- 
try, which  finds  its  practical  applications  in  architecture  and  machine 
drawing.  From  geometrical  statics  we  obtain  also  a  construction  method 
or  routina — viz.,  graphical  statics — which  finds  its  practical  applications  in 
the  graphical  calculation  of  structures  and  machines.  Both  descriptive 
geometry  and  graphical  statics  have  still,  with  reference  to  these  practical 
ends,  to  develop  and  make  use  of  the  general  relations  which  subsist  be- 
tween the  geometrical  constructions  to  which  they  give  rise,  and  thus  each, 
according  to  its  means,  contribute  to  the  discovery  and  spread  of  geo- 
metrical and  mechanical  truths. 

From  this  co-ordination  of  descriptive  geometry  and  graphical  statics 
we  must  not,  however,  infer  an  equal  importance  ;  for,  while  in  geometri- 
cal drawing  we  have  always  to  represent  an  ideal  image,  and  the  graphical 
method  is  therefore  directly  suggested,  we  have  for  statical  calculations 
the  analytical  process  also  at  our  disposal,  and  everything  depends  then 
upon  the  relative  advantages  and  disadvantages  of  the  graphical  and  ana- 
lytical methods.  We  have  thus  noticed  all  the  most  important  points 
which  occur  in  a  theoretical  consideration,  and  there  only,  remains  to  make 
a  comparison  from  a  practical  standpoint  (X.). 


VIII. 

THE  MODERN  GEOMETRY. 

Geometry  treats  of  figures  or  constructions  in  space.  These  figures  and 
their  properties  are  not  always  regarded  and  treated  in  equal  extent  and 
generality. 

Geometrical  knowledge  found  its  origin  in  practical  needs,  and  the 
ancients  confined  themselves  almost  exclusively  to  special  investigations 
of  individual  figures  and  bodies  of  definite  form,  such  as  presented  them- 
selves to  the  eye.  In  the  pTiorisms  of  Euclid  (-285),  according  to  Pappus 
(end  of  the  fourth  century),  the  mutual  relations  of  the  circle  and  straight 
lines  were,  indeed,  given  with  a  certain  degree  of  completeness,  but  these 
have  not  come  down  to  us. 

Properties  thus  determined  had  naturally  only  a  limited  significance, 
and  could  neither  count  upon  permanence  nor  give  satisfactory  conclu- 
sions. Investigators  sought,  therefore,  assistance  where  it  was  best  afforded, 
in  analysis.  This  was,  in  the  sixteenth  century,  by  the  algebra  of  Vieta 
(1540-1603),  notably  enriched. 


Xl  THE    MODERN    GEOMETRY.  [iNTROD. 

From  this  period  geometry,  for  a  long  time,  served  merely  as  an  aid  to 
analysis,  interpreting  graphically  its  results  (V.).  From  this  union  the 
greatest  advantages  were  derived,  as  analysis  led  to  the  infinitesimal  cal- 
culus of  Newton  and  Leibnitz,  and  geometry  to  the  analytical  geometry  of 
Descartes  (1596-1650). 

But  the  extension  and  generality  which  geometrical  truths  received  by 
this  great  creation  of  Descartes  was  essentially  due  to  analysis.  Desaryucts 
(1593-1662)  and  Pascal  (1623-1662)  extended  pure  geometrical  considera- 
tions, and  made  the  first  step  towards  the  modern  geometry  when  they 
regarded  the  conic  sections  as  projections  of  the  circle,  and  deduced  the 
properties  of  the  first  from  those  of  the  last.  Then  De  la  Hire  (1640-1718), 
Le  Poivre  (1704)  and  Hnygens  (1629-1695)  occupied  themselves  with  geo- 
metrical investigations.  While  the  two  first  developed  the  methods  of 
Desargues  and  Pascal,  Huygens  and,  later,  Newton  (1642-1727)  applied 
pure  geometry  in  optics  and  mechanics.  Soon,  however,  the  Calculus  of 
Newton  and  Leibnitz  (1684  and  1687)  showed  itself  so  wonderfully  fertile 
in  analytical  geometry,  that  geometry  proper  was  put  in  the  background. 
Only  a  few,  as  Lambert  (1728-1777),  still  regarded  it  with  favor. 

Then  appeared  Monge  (1728-1777),  and  gave  the  impulse  to  a  complete 
revolution  in  geometrical  views,  and  to  the  reconstruction  of  the  science 
upon  a  new  basis.  In  his  Lemons  de  Geometric  descriptive  (Paris,  1788),  all 
those  problems  previously  treated  in  a  special  and  uncertain  manner  in 
stereotomy,  perspective,  gnomonics,  etc.,  were  referred  back  to  a  few  gen- 
eral principles,  and,  without  the  aid  of  analysis,  the  most  important  prop- 
erties of  lines  and  surfaces  were  deduced.  While  descriptive  geometry 
taught  the  relations  between  bodies  in  space  and  drawn  figures,  it  strength- 
ened the  power  of  abstraction  ;  introducing  into  geometry  the  transforma- 
tion of  figures,  it  gave  to  its  deductions  an  advantage  till  then  possessed 
only  by  analysis;  and  while,  finally,  it  owed  its  comprehensive  results  to 
the  application  of  projections,  it  pointed  the  way  for  the  further  develop- 
ment of  geometry  itself. 

Meanwhile,  in  the  field  of  analytical  geometry,  the  conclusion  had  been 
reached  that  the  desired  truths  admitted  of  a  still  more  general  compre- 
hension. All  properties  had  been  obtained  only  with  respect  to  and  by 
means  of  a  determinate  co-ordinate  system.  But  already  Godin  (1704- 
1760)  had  announced  "  que  I" art  de  decouvrir  les  proprietes  dss  courses  est 
a  proprement  parler,  I1  art  de  changer  le  systeme  de  co-ordonnees"  (Traite  den 
proprietes  communes  a  toutes  les  courses).  This  idea  Car  not  seized  upon 
(1753-1823),  and  in  the  sixth  chapter  of  his  Geometric  de  position  (Paris, 
1803)  he  sought  to  obtain  a  more  general  comprehension  of  figures  by 
analysis,  and  to  avoid  the  indeterminancy  of  this  last  by  the  introduction 
of  the  idea  of  position,  and  by  many  solutions  after  the  method  already 
pointed  out  by  Liebnitz  and  d'Alembert. 

Now  began  a  veritable  race  in  the  condensation  and  promulgation  of 
geometrical  truths,  in  which  the  pure  geometrical  method  obtained  the 
palm.  The  scholars  of  Monge — Brianchon,  Servois,  Chasles,  Poncelet,  Ger- 
gonne — working  with  him  and  in  his  spirit,  filled  the  Annales  des  mathe- 
matiyues  and  the  Correspondance  sur  Vecole  polytechnique  with  new  re- 


INTBOD.]  THE   MODERN    GEOMETKY.  xli 

suits — the  two  last  named  discovering  the  general  law  of  reciprocity  or  du- 
ality. 

The  foundation  proper  of  the  modern  geometry  was 
his  Traite  des  proprietes  projectiles  des  figures  (Paris,  l828fi"Aggrandir  les 
resources  de  la  simple  Geometrie,  en  generaliser  les  conceptions  et  le  langage  or- 
dinairement  assez  restreints,  lesrapproclier  deceuxde  la  Geometrie  analytique, 
et  surtout  offrir  des  moyens  generaux,  propres  a  demontrer  et  a  fair  e  decouvrir, 
d'une  maniere  facile,  cette  classe  de  proprietes  dont  jouissent  les  figures  quand 
on  les  considere  d'une  maniere  purem'ent  abstraite  et  independamment  d'au- 
cune  grandeur  absolue  et  determinee,  tel  est  Vobjet  qu'on  s'est  speciolemejit 
propose  dans  cet  ouvrage."1"1 

The  new  ideas  found  in  Germany  especially  fruitful  soil.  Mobius, 
Pliicker,  Steiner,  Grassman,  and  many  others,  proceeding  in  part  from 
entirely  different  points  of  view,  opened  out  an  abundance  of  new  direc- 
tions which  have  not  yet  been  thoroughly  explored,  and  which,  in  union 
with  other  investigations,  have  caused  a  thorough  change  in  our  concep- 
tions of  space  relations,  whose  latest  phases  are  indicated  by  the  names  of 
Riemann,  Helmlioltz  and  Lie-klein. 

In  this  development  period,  also,  still  existed  the  two  parties  in  analyti- 
cal and  synthetic,  or  pure  geometry.  Plucker  held  the  analytical  relations 
as  the  most  general,  and  which  were  with  advantage  to  be  illustrated  and 
interpreted  geometrically  ;  while  Steiner  recognized  in  the  space  figure 
itself  the  true  object  and  most  efficient  aid  of  investigation.  Both  direc- 
tions— the  modern  analytic  and  synthetic — lead  naturally  to  the  same  results. 
"With  reference  to  the  methods,  however,  they  diverge  the  nearer  the  ideas 
and  transformations  of  geometry  approach  the  generality  and  ease  of  the 
alg3braic  method,  thus  rendering  possible  an  abandonment  of  this  last. 
Thus,  while  analytical  geometry,  through  the  theory  of  determinants  of 
Hesse,  came  into  ever  closer  connection  with  analysis — a  direction  in  which 
English  and  Italian  investigators — as  Salmon,  Cayley,  Cremona — brilliantly 
assisted,  the  Erlangen  Professor  von  Staudt  cut  loose  from  algebraic  formu- 
la and  metrical  relations,  and  gave  us  the  geometry  of  position  (Nilrnberg, 
1847,  Beitr.  z.  Geom.  d.  Lage). 

After  von  Staudt,  the  strict  geometry  of  position  remained  a  long  time 
disregarded,  while  the  synthetic  geometry  of  Steiner  has  enjoyed,  without 
intermission  till  the  present  day,  a  special  preference  on  the  part  of  mathe- 
maticians. One  reason  may  indeed  be  that  mathematicians  take  little  in- 
terest in  an  independence  of  geometry  to  which  analysis  can  lay  no  claim ; 
but  another,  still  more  potent,  is  the  extremely  condensed,  almost  schematic 
presentation  of  von  Staudt,  which  has  not  exactly  an  encouraging  effect 
upon  every  one. 

Culmann  gave  the  impulse  to  a  change  in  this  respect.  In  his  graphical 
statics  he  rests  directly  upon  the  work  of  von  Staudt,  and,  with  something 
more  than  boldness,  assumes  a  knowledge  of  the  geometry  of  position 
among  all  practical  men.  Such  a  course  was  not  indispensable  for  the 
foundation  of  his  method,  and  impeded  the  spread  of  the  graphical  stat- 
ics ;  but  by  it  the  geometry  of  position  gained.  This  last  had  next,  of 
necessity,  to  be  introduced  into  the  Zurich  Polytechnic,  and  thus' arose  the 


MODERN  GEOMETRY  IN  ENGINEERING  PRACTICE.      [iNTROD. 

first,  until  now,  only  complete  text-book  upon  the  subject,  the  "  Geometric 
der  Lage,"  by  Reye  (Hannover,  1868),  as  the  direct  result  of  the  graphical 
statics  of  Culmaun. 

Since  then,  the  modern  geometry  has  been  introduced  into  all  technical 
institutions  throughout  Germany,  and  thus  placed  at  the  disposal  of  the 
arts  and  sciences. 

As,  according  to  its  founder,  Poncelet,  it  reaches  the  highest  range  of 
speculation,  so  also  in  the  most  practical  relations  it  acts  to  simplify  and 
condense :  "  Pen  d  peu  les  connaissances  algebriques  deviendront  mains  in- 
dispensdbles,  et  la  science,  reduite  d  ce  qu'elle  doit  etre,  d  ce  quSelle  devrait  etre 
dejd,  sera  aimi  mise  d  la  portee  de  cette  classe  d' homines,  qui  n'a  que  des  mo- 
ments fort  rares  d  y  consacrer." 

[For  illustrations  of  the  method  of  the  modern  geometry,  the  reader  may 
consult  the  Appendix  to  this  chapter.] 


IX. 

THE   MODERN    GEOMETRY    IN   ENGINEERING   PRACTICE. 

One  who  should  infer  that  a  science  created  thus  from  its  very  inception 
with  reference  to  the  needs  of  practice*  must  have  found  access,  above  all, 
in  technical  circles,  would  be  much  mistaken.  As  Culmann  sent  out  his 
graphical  statics,  deep  silence  prevailed,  and  if  the  modern  geometry  ap- 
peared here  and  there  in  the  lecture  plan  of  one  and  another  polytechnic,  it 
was,  without  doubt,  due  to  the  zeal  of  some  enthusiastic  privat  docent  who 
had  undertaken  the  thankless  task  of  holding  forth  to  empty  benches. 

Whence  came  this  indifference  to  a  discipline  proceeding  from  the  Ecole 
polytechnique  ?  It  is  hard,  indeed,  to  find  a  sufficient  reason.  We  often 
hear  it  said  that  by  reason  of  the  colossal  extension  which  engineering 
sciences  have  experienced,  students  are  already  overburdened.  Most  true  ! 
and  it  is  just  here  that  the  modern  geometry  comes  to  our  assistance.  It 
is  precisely  to  this  that  the  learned  critic  of  Monge,  Dupin,  alludes  :  "  II 
(terrible  que  dans  Vetat  actuel  des  sciences  mathematiques  le  seul  moyen  d'ern- 
pecher  que  leur  domaine  ne  devienne  trop  vaste  pour  notre  intelligence,  c'est 
de  generaliser  de  plus  en  plus  les  theories  que  ces  sciences  embrassent,  afin 
qu'un  petit  nombre  des  verites  generates  et  fecondes  soit  dans  la  tete  des 
Tiommes  I1  expression  abregee  de  la  plus  grande  variete  des  f aits  particuliers" 

The  modern  geometry  in  its  present  form  starts  with  a  small  number  of 
elementary  constructions  whose  properties  are  first  set  forth,  and  then,  pro- 
ceeding from  these  by  combination  and  comparison,  it  covers  the  entire 
department  of  space.  The  engineer,  during  and  after  his  preparation,  has 
to  do  with  space  problems,  with  geometrical  principles  and  constructions ; 

*  Poncelet  himself  set  upon  the  title-page  of  his  work  :  "  Ouvrage  utile  d 
ceux  qui  s^occupent  des  applications  de  la  Geometric  descriptive  et  d' operations 
geometriques  sur  le  terrain." 


INTBOD.]       PRACTICAL  SIGNIFICANCE  OF  GEAPHICAL  STATICS.  xliii 

"how  many  superfluous  definitions  and  demonstrations  could  not  be 
spared,  if  they  were  already  completely  comprehended  and  recognized  by 
the  scholar  as  parts  of  a  higher  whole"  (Gulmann — "Die  Graphische 
Statik  ").  At  no  very  distant  day  it  will  no*  longer  be  possible  to  read  a 
scientific  work  upon  applied  mathematics  without  familiarity  with-  the 
principles  of  the  modern  geometry.*  Permitting  pure  graphical  applica- 
tions, without  the  aid  of  analytic  symbols,  it  forms  the  common  point  of 
view  for  descriptive  geometry,  practical  geometry,  and  graphical  statics. 

Descriptive  geometry  existed  before  the  modern,  and  this  last  has  sprung 
from  it.  Now,  reversely,  the  geometry  of  position  comes  to  the  aid  of 
descriptive  geometry,  and  offers  in  return  its  most  fruitful  principles  and 
efficient  aid.  Thus  in  descriptive  geometry  we  may  refer  to  the  works  of 
Pohlke,  tichlesinger,  and  Fiedler.  The  effect  of  the  geometry  of  position  in 
this  direction  to  simplify  and  condense  may  be  seen  from  the  work  of 
Staudigl  ("Ueber  die  Identitat  von  Constructionen  in  perspective,  schiefer 
und  orthogonaler  Projection'1),  where  it  is  proved  that  "all  problems  of 
the  descriptive  geometry,  in  which  neither  linear  nor  angular  measure  are 
considered — therefore  all  problems  which  belong  to  the  geometry  of  posi- 
tion— can  in  similar  manner  and  by  precisely  similar  constructions  be  solved 
as  well  in  perspective  as  in  oblique  and  orthagonal  projection."  In  shades 
and  shadows  and  in  geometrical  drawing,  Burmeister  and  Paulus  owe  to 
the  modern  geometry  the  simplicity  of  their  constructions. 

In  the  department  of  practical  geometry  also,  in  geodesy,  perspective, 
surveying,  we  mark  the  influence  of  the  modern  geometry  in  the  works  of 
Mutter  and  Spangeriberg,  of  Franke  and  Baur. 

In  mechanics  and  physics,  we  see  it  again  in  the  works  of  Lindemann, 
Burmeiater  and  Zech. 


PEACTICAL  SIGNIFICANCE  OF  THE  GBAPHICAL  STATICS. 

We  have  already  remarked  (VII.)  that  the  importance  of  graphical 
statics  is  in  great  part  dependent  upon  its  advantages  as  compared  with 
the  analytical  method,  and  have  reserved  for  this  place  a  comparison  from 
a  practical  point  of  view. 

Here,  first  of  all,  we  have  to  notice  the  independence  of  the  graphical 
construction  of  the  regularity  or  irregularity  of  the  given  relations. 
Whether  the  forces  are  equal  or  not,  whether  they  act  at  equal  or  varying 
distances,  even  their  relative  position,  are  matters  of  indifference.  Centre 
of  gravity,  central  ellipse,  kernel — for  all,  even  the  most  irregular  figures, 
are  found  in  similar  manner,  with  equal  ease,  even  when  exact  analytical 
solutions  are  hardly  conceivable.  Thus  a  process,  a  routine  almost 
mechanical  is  rendered  possible  in  many  investigations  of  stability,  with- 
out losing  sight  of  interior  relations ;  for  in  the  repeated  and  independent 
compositions  of  the  forces  we  always  perceive  the  origin,  connection  and 

*  Well  illustrated  in  Gillespie's  Land  Surveying.     New  York,  1870. 


PRACTICAL  SIGNIFICANCE  OF  GRAPHICAL  STATICS.      [iNTKCD. 

reason  of  the  result  obtained,  which,  in  the  substitution  of  numbers  in  a 
formula,  is  not  always  the  case. 

With  this  advantage  goes  hand  in  hand  a  disadvantage.  This  very 
regularity  of  the  process  is  a  consequence  of  its  special,  we  might  almost 
say  numerical,  character  (I.).  In  a  numerical  analytical  example  greater  or 
less  regularity  has  also  but  little  effect.  This  numerical  character  has  also 
for  consequence  that  we  can  never  attain  to  general  laws  and  relations 
(IV.,  VII). 

The  practical  engineer  becomes  with  time  ever  more  familiar  with  the 
dividers  and  rule,  while  facility  in  analytical  operations  gradually  disap- 
pears. A  graphical  construction  once  completed  is  not  easily  forgotten,  or 
a  single  glance  at  a  similar  figure  suffices  to  recall  the  whole  process.  It  is 
indeed  easy  in  clearly  given  formulae  to  substitute  special  numerical  values ; 
but  formulas  unfortunately  are  not  always  clearly  given,  in  some  cases  can- 
not be  so  given,  without  presuming  upon  the  thorough  familiarity  of  the 
reader  with  the  processes  involved ;  these  and  the  very  many  and  various 
systems  of  notation  in  use  leave  to  the  constant,  easily  acquired  and 
remembered  graphical  solution  many  advantages. 

But  here  we  may  remark  that  graphical  solutions  can  only  be  easily 
acquired,  retained  or  quickly  recovered  when  the  constructions  are  based 
upon  methods  purely  geometric,  and  not  when  they  are  simply  the  interpre- 
tation of  previously  obtained  analytical  results.  In  the  latter  case  we 
must  recall  the  process  of  development  of  the  formula  as  well  as  the 
graphical  construction,  and  the  method  is  thus  too  often  confusing  instead 
of  simple. 

Often  it  is  desired  to  make  visible  the  results  of  an  investigation,  as  in 
the  case  of  the  arch,  where  the  graphical  method  is  especially  advan- 
tageous, and  has  in  France  been  long  used  (VII.). 

Errors  relating  to  the  mutual  relation  of  strains  are  more  easily  discov- 
ered in  graphical  solutions  than  in  analytical,  as  a  certain  law  of  regularity 
is  always  visible,  which  breaks  abruptly  for  an  error  in  construction.  By 
calculation,  on  the  other  hand,  we  can  more  easily  select  any  one  place  in 
the  structure,  and  determine  the  strain  there  independently  of  the  others. 

As  to  which  of  the  two  methods  demands  the  least  time  is  a  matter  of 
minor  importance.  In  a  construction  costing  from  thousands  to  millions, 
it  matters  little  whether  the  calculations  require  one  or  several  days,  more 
or  less,  if  only  the  results  are  clear  and  correct.  It  is  a  question  also 
which  can  hardly  be  decided  in  favor  of  one  or  the  other,  dependent  as  it 
is  upon  elements  other  than  those  pertaining  to  the  methods  themselves — 
such  as  varying  individual  skill  and  capacity  in  either  direction.  The 
declaration  which  is  already  sometimes  encountered,  that  the  numerical 
calculation  of  a  continuous  girder  requires  about  three  times  as  much  time 
as  the  graphical  solution,  sounds  questionable.  Why  not  at  once  furnish 
the  statement  with  decimal  places  f  In  general,  for  ordinary  cases,  the  ana- 
lytical solution  requires  less  time ;  for  irregular  and  more  complicated  cases, 
the  graphical. 

The  exactness  of  the  graphical  solution  is  sufficient,  but  it,  too,  depends 
upon  the  care  and  skill  of  the  draughtsman.  The  greater  the  forces  and 


IXTEOD.]  LITERATURE   UPON    GRAPHICAL    STATICS.  xlv 

dimensions  with  which  one  works,  the  better  the  results  obtained.  The 
scales  should  not,  then,  be  taken  too  small. 

It  is  hoped  that  these  considerations,  now  drawing  to  a  close,  will  suffice 
to  give  the  reader  clear  ideas  upon  the  nature  and  origin,  advantages  and 
disadvantages,  of  the  graphical  statics.  The  determination  whether  he  will 
enter  more  fully  into  the  subject — it  may  be,  even  take  part  in  its  develop- 
ment (there  is  abundance  of  room  for  workers),  and  in  this  case  the  choice 
of  direction  may  thus  be  facilitated. 

The  graphical  statics  is  certainly  suited,  especially  in  extended  applica- 
tions of  the  geometry  of  position,  to  furnish  many  new  points  of  view,  and 
in  a  practical  respect  it  can  often  greatly  simplify.  Whoever  has  really 
studied  the  new  methods  must  admit  this. 

On  the  other  hand,  the  importance  of  the  graphical  statics  is  sometimes 
exaggerated.  It  appears  out  of  place  when  in  works  designed  for  practice 
graphical  solutions  are  given  of  problems  which  any  reasoning  being  can 
almost  solve  in  his  head. 

Such  solutions  may  find  a  place  in  special  text-books  upon  the  subject, 
where  they  may,  indeed,  be  desirable  for  completeness. 

If  it  is  desired  to  make  two  independent  investigations  of  stability,  as 
for  large  and  important  constructions  is  always  desirable,  it  will  be  found 
of  advantage,  if  a  suitable  graphical  solution  exists,  to  make  the  first  deter- 
mination graphically.  Nothing  more  ensures  a  conviction  of  the  correct- 
ness of  an  investigation  than  a  correspondence  of  the  graphical  and  cal- 
culated results. 


XI. 

LITERATURE   UPON   GRAPHICAL    STATICS. 

We  have  already  referred  in  VI.  to  the  most  important  contributions  in 
the  branch  of  graphical  statics,  and  now  annex  a  list  of  the  literature  upon 
the  subject  so  far  as  known  to  us. 

Where  several  works  treat  of  the  same  subject,  we  have  allowed  ourselves 
a  brief  critical  notice.  Opportunity  is  thus  given  to  those  who  would  take 
part  in  the  development  of  graphical  statics  to  make  themselves  acquainted 
with  all  existing  works,  and  at  the  same  time  the  practical  man  is  enabled 
in  any  case  that  may  come  up  to  inform  himself  as  to  where  assistance 
may  best  be  sought.  A  short  remark  to  specify  the  contents  may  in  this 
respect  often  help  in  the  right  direction.  The  succession  is  in  each  division 
chronologically  arranged 

Although  the  literature  of  the  subject  would  seem  from  the  following 
tolerably  extensive,  still  the  number  of  pure  geometrical  solutions  in 
which  no  analytical  formulae  appear  is  much  less.  Publications  upon  the 
subject  would,  moreover,  beyond  doubt,  be  still  more  numerous  were  it 
not  for  the  difficulty  and  cost  of  production  of  lithograph  plates. 


LITEKATUliE    UPON    GRAPHICAL    STATICS. 


I.   TEXT-BOOKS  UPON  GRAPHICAL   STATICS. 

Culmann,  K.—"  Die  graphische  Statik."  With  Atlas  of  36  Plates.  Zurich, 
Meyer  and  Zeller,  1866.  [I.  Part,  1864:  Elements  and  Graphical 
Investigations  of  Structures.  Also,  second  edition,  first  volume, 
1875,  with  17  Plates.  General  Principles,  second  volume,  to  follow 
shortly.] 

Bauschinger — "Elemente  der  graphischen  Statik."  With  Atlas  of  20 
Plates.  Miinchen,  1871.  [Without  the  aid  of  modern  geometry,  and 
without  practical  applications.  Admirable  exposition  of  the  Princi- 
ples.] 

lleuleaux. — An  outline  of  the  graphical  statics  is  to  be  found  in  "  Der  Con- 
structeur,"  by  Reuleaux,  third  ed.  Braunschweig,  1872. 

Levy — "  La  Statique  Graphique  et  ses  Applications."  Paris,  1874.  With 
Atlas  of  24  Plates.  [Principles  and  several  applications;  clear  and 
elegant  exposition  of  the  subject.] 


n.   PAPERS  UPON  THE   GRAPHICAL   STATICS. 

Most — "  Ueber  eine  allgemeine  Methode,  geometrisch  den  Schwerpunkt 
beliebiger  Polygone  und  Polyeder  zu  bestimmen."  Archiv  d.  Math, 
und  Phys.,  IL.  (1869),  p.  355.  [Also  applicable  to  curve  areas,  with- 
out equilibrium  polygon.] 

Culmann,  K. — "Ueber  das  Parallelogram  und  iiber  die  Znsammensetzung 
der  Krafte."  Vierteljahrsschr.  d.  Naturforsch.  Ges.  zu  Zurich,  1870. 
[Correspondence  of  the  graphical  statics  with  the  Statics  of  Pliicker.] 

Mohr — "  Bsitrag  zur  Theorie  der  Holz-  und  Eisenconstructionen."  Zeitschr. 
d.  Hannov.  Arch.  u.  Ing.  Ver.,  1870,  p.  41.  [Relation  between  the 
neutral  axis  and  centre  of  strains.] 

Grunert,  J..A. — "Ueber  eine  Graphische  Methode  zur  Bestimmung  des 
Schwerpunktes  eines  beliebigen  Vierecks."  Arch.  d.  Math.  u.  Phys., 
LII.  (1871),  p.  494.  [Simple  and  brief.  Compare  also  L.,  p.  212.] 

Cremona,  B. — "  Le  figure  reciproche  nelle  statica  grafica."  With  5  Plates. 
Milan,  1872.  German  translation  in  Zeitschr.  d.  Ost.  Arch.  u.  Ing. 
Ver.,  1873,  p.  230.  [Force  and  equilibrium  polygon  as  reciprocal 
figures.] 

Du  Bois,  A.  J.— "  The  New  Method  of  Graphical  Statics."  Van  Nostrand's 
"Engineering  Magazine,"  Vol.  XII.,  Nos.  74,  75,  76,  77,  78.  [General 
properties  of  force  and  equilibrium  polygons,  with  practical  applica- 
tions to  bending  moments,  and  several  important  mechanical  problems. 
Also,  Maxwell's  Method  applied  to  bridges,  roof  trusses,  etc.]  Sepa- 
rate reprint,  1875.  Van  Nostrand,  New  York. 


LITERATURE    UPON   GRAPHICAL    STATICS. 


III.  APPLICATION  TO  THE   SIMPLE   GIRDER. 

Culmann,  K.— "Der  Balken."  Third  chap,  of  d.  graph.  Statik,  1866. 
[Contains  also  the  construction  of  the  inner  forces.]  I 

Vojdcek — "  Graphische  Bestimmung  der  Biegungsmomente  an  kurzen 
Tragern."  Zeitschr.  d.  Vereins  Deutsch.  Ing.,  1868,  p.  503.  [Graphi- 
cal interpretation  of  analytical  relations.] 

Cotterill,  J.  H. — "On  the  Graphic  Construction  of  Bending  Moments." 
"Engineering,"  1869  (VII.),  p.  32.  [Equilibrium polygon  for  the  sim- 
ple truss,  with  references  to  Rouleaux  and  Culmann.] 

Winkler,  E. — "Einfache  Trager,"  "Theorie  der  Brticken,"  "Aeussere 
Krafte  gerade  Trager."  Wien,  1872.  [Simultaneous  presentation  of 
analytical  and  graphical  methods.] 

Ott,  K.  von — "Wirkung  paralleler  Krafte  auf  einfache  Trager  mit  Gerade 
Langenachse."  In  die  Grundziige  d.  graph.  Reclmens  u.  d.  graph. 
Statik.  Prag,  1872.  p.  28.  [The  most  elementary  principles  pertain- 
ing to  composition  of  forces  in  a  plane  are  prefaced.] 

IV.  APPLICATION  TO   THE   CONTINUOUS  GIRDER, 

Culmann,  K. — "Der  continuirliche  Balken."  Fourth  chap,  of  the  Graph. 
Statik,  1866.  [With  examples — the  moments  at  the  supports  are 
analytically  determined.] 

MoJir — "  Beitrag  zur  Theorie  der  Holz-  und  Eisehconstructionen."  Zeitschr. 
d.  hannov.  Arch.  u.  Ing.  Ver.,  1868,  p.  19.  [Completion  of  Culmann' s 
method — the  moments  at  the  supports  are  graphically  determined.] 

Lippich — "Theorie  des  continuirlichen  Tragers  Constanten  Querschnitts." 
Wien,  1871.  Separate  reprint  from  Forster's  Bauzeit.,  1871,  p.  103. 
[Graphical  method,  together  with  elementary  analytical.] 

Hitter,  W. — "  Die  elastische  Linie  und  ihre  Anwendung  auf  den  continuir- 
lichen Balken."  Zurich,  1871.  [Mohr's  method— given  as  a  supple- 
ment to  the  Graph.  Statik  of  Culmann.] 

Winkler,  E. — "  Continuirliche  Trager.  Theorie  der  Bracken — aeussere 
Krafte  gerade  Trager."  Wien,  1872.  [The  Mohr-Culmann  method, 
together  with  analytical.] 

Solin,  J. — "  Geometrische  Theorie  der  continuirlichen  Trager."  Mitth.  d. 
Arch.  u.  Ing.  Ver.  in  Bohmen,  1873. 

Greene,  Chas.  E. — "  Graphical  method  for  the  analysis  of  Bridge  Trusses  ; 
extended  to  Continuous  Girders  and  Draw  Spans."  New  York,  1875. 
[Moments  at  supports  found  by  successive  approximation,  or  balancing 
of  moment  areas.] 

V.  APPLICATION  TO  FRAME  WORK. 

Culmann,  K.—"  Das  Fachwerk."  Fifth  chap.  Graph.  Statik,  1866.  [Most 
general  form  of  parallel  truss,  suspension  truss,  Pauli's  truss,  roof 
trusses.] 


LITERATURE    UPON    GRAPHICAL    STATICS. 

Keck,  W.— "  Ueber  die  Erinittclung  der  Spannungen  in  Fachwerk  tragern 
mit  Hulfe  der  graphischen  Statik."  Zeitsclir.  d.  hannov.  Arch.  u.  Ing. 
Ver.,  1870,  p.  153.  Separate  reprint,  Hannover,  1872.  [Presentation 
of  the  method  with  reference  to  practice.] 

Jenkin — "  On  the  Practical  Application  of  Reciprocal  Figures  to  the  Cal- 
culation of  Strains  in  Frame-work.  Transact,  of  the  R.  Soc.  of  Edin- 
burgh, 1870,  (XXV.)  p.  441. 

Maxwell,  Prof.  Cleric — "  Reciprocal  Figures,  Frames,  and  Diagrams  of 
Forces."  Trans,  of  R.  Soc.  of  Edinburgh,  1869-70. 

Uhwin — "  Iron  Bridges  and  Roofs.'1  London,  1869.  [Application  to  roof 
trusses,  wind  force,  etc.] 

Banfon,F.A.—"The  Strains  in  Trusses."  New  York,  Appleton,  1872. 
[Examples  of  simple  trusses  drawn  to  scale.] 

Bow,  Robert  H. — "Economics  of  Construction  in  Relation  to  Framed  Struc- 
tures." London,  1873.  [Application  of  Maxwell's  Method  only  to 
roof  trusses,  etc.] 

Ott,  K.  von — "  Das  Fachwerk."  In  Grundzuge  d.  graph.  Rechnens  u.  d. 
graph.  Statik.  Prag,  1872.  [Roof  trusses,  truss  fixed  at  one  end  and 
free  at  the  other,  bridge  trusses.] 

Reuleaux — "  Hilfslehren  aus  der  Grapho  statik."  Second  chap,  of  the  Con- 
structeur,  third  ed.,  1872.  [Compound  truss,  roof  trusses,  etc.] 

Scliaffer — "Graphische  Ermittelung  der  Ordinaten  des  Schwedler'schen 
Tragers."  Zeitschr.  fur  Bauwesen,  1873,  p.  237.  [Proceeding  from 
the  equation  for  the  same.] 

Heuser — "Graphische  Ermittelung  der  Ordinaten  des  Schwedler'schen 
Tragers."  Zeitschr.  f.  Bauwesen,  1873,  p.  523.  [Preceding  method 
simplified — another  by  means  of  equilibrium  polygon.] 

VI.  APPLICATION  TO  THE  IRON  ARCH. 

Culmann,  K.—"  DerBogen."  Sixth  chap,  der  graph.  Statik,  1866.  [Con- 
tains also  the  inverted  or  suspended  arch.  The  arch  as  a  rigid  body.] 

Durand-Claye,  A. — "  Sur  la  verification  de  la  stabilite  des  arcs  m&alliqucs 
et  sur  1'emploi  des  courbes  de  pression."  Ann.  d.  ponts  et  chauss., 
1868,  sem.  1,  p.  109.  [Mery-Durand  pressure  curves,  but  with  refer- 
ence to  the  absolute  resistance  of  the  material.] 

Frankel,  W. — "  Zur  Theorie  der  elastischen  Bogentrager."  Zeitschr.  d.  han- 
nov. Arch.  u.  Ing.  Ver.,  1869,  p.  115.  [Following  out  Mohr's  idea  of 
the  equilibrium  polygon  as  elastic  line.  ] 

MoJir — "  Beitrag  zur  Theorie  der  elastischen  Bogentrager."  Zeitschr.  d. 
hannov.  Arch.  u.  Ing.  Ver.,  1870,  p.  389.  [Criticism  of  the  preceding 
method,  and  giving  another.] 

Beitrage  zur  graphischen  Berechming  elastischer  Bogentrager  mit 
Kampfergelenken."     Mitth.  d.  Arch.  u.  Ing.  Ver.,  in  Bohmen,  1873. 


LITERATURE    UPON    GRAPHICAL   STATICS. 


VII.  APPLICATION  TO  THE  ARCH. 

Cousinery,  E.  B. — "  Application  des  proce'de's  du  calcul  graphique  a"  divers 
problemes  de  stability."  Fourth  chap,  of  Calcul  par  le  Trait.  Paris, 
1839.  [With  especial  reference  to  the  strength  of  abutments — pure 
graphical  treatment.] 

M&ry — "  MSmoire  sur  I'Squilibre  des  voutes  en  berceau."  Ann.  d.  ponts 
et  chauss.,  1840,  sem.  1,  p.  50.  [Geometrical  determination  of  every 
possible  pressure  curve.] 

Culmann,  K. — "  Der  Bogen.".  Sixth  chap,  of  Graph.  Statik,  1866.  [Con- 
taining also  arch  centerings  ;  exact  discrimination  of  support  and 
pressure  line.] 

Durand-Claye,  A. — "Sur  la  verification  de  la  stability  des  voutes  en 
maconnerie  et  sur  1'einploi  des  courbes  de  pression."  Ann.  d.  ponts 
et  chauss.,  1867,  sem.  1,  p.  63.  [Reference  to  relative  resistance  of 
material.  ] 

Harlacher,  A.  R.— "Die  Stiitzlinie  im  Gewolbe."  Tech.  Blatter,  1870,  p 
49.  [Practical  method  by  inscription  of  support  line,  according  to 
Culmann.] 

Heuser — "Zur  Stabilitatsuntersuchung  der  Gewolbe."  Deutsche  Bauzeit. 
1872,  p.  365.  [Also  methods  for  unsymmetrical  form  and  load.] 


VHI.   APPLICATION  TO  RETAINING  WALLS. 

Poncelet.  J.  V. — "  Me'moire  sur  la  stabilite  des  reve'tements  et  leur  fonda- 
tion."  Mem.  de  1'off.  du  Ggnie,  1838  (XIII.) ;  separate  reprint, 
Paris,  1840.  [First  analytical  graphical  theorie.] 

Cousinery,  E.  B. — "  Application  des  precedes  du  calcul  graphique  &  divers 
problemes  de  stability."  Hourth  chap,  of  "  Calcul  par  le  Trait,"  1839. 
[Pure  graphical,  without  formulae.] 

^aint-G-uilhem — "  MSmoire  sur  la  poussSe  des  terres  avec  ou  sans  sur- 
charge." Ann.  d.  ponts  et  chauss.,  1858,  sem.  1,  p.  319.  [Further 
development  and  generalization  of  Poncelet's  Theory.] 

Rarikine — "Manual  of  Civil  Engineering."  London,  fourth  ed.,  1865. 
[Containing  graphical  construction  of  pressure  parallel  to  earth  sur- 
face upon  vertical  wall.]  i 

Culmann,  K. — "  Theorie  der  Stutz-  und  Futter-Mauern."  Eighth  chap,  of 
Graph.  Statik,  1866.  [With  use  of  equilibrium  polygon,  pressure 
upon  tunnel  arches.] 

JMzhey,  E. — "  Beitrage  zur  Theorie  des  Erddrucks  und  graphische  Bestim- 
mung  der  Sta'rke  von  Futter-Mauern."  Mitth.  iiber  Gegenst.  d.  Artill. 
und  Geniewesens;  separate  reprint,  with  two  plates,  Wien,  1871. 
[Point  of  application  of  earth  pressure  for  complicated  contour.] 

MoJir — "Beitrage  zur  Theorie  des  Erddrucks."  Zeitschr.  des  hannov. 
Arch.  u.  Ing.  Ver.,  1871,  p.  344.  [Point  of  application  of  earth  pres- 
sure and  new  analytical  theory.] 


1  GRAPHICAL   DYNAMICS. 

WinTder,  K-"  Neue  Theorie  des  Erddrucks."  Wien,  1872.  [Containing 
graphical  methods  according  to  the  old  theory.] 

Haseler,  0. — "  Beitrage  zur  Theorie  der  Futter-  tmd  Stiitz-Mauern."  Zeit- 
schr.  d.  hannov.  Arch.  u.  Ing.  Ver.,  1873,  p.  36.  [Graphical  deter- 
mination of  earth  pressure  according  to  Culmann.] 

MISCELLANEOUS  APPLICATIONS. 

Reuleaux—"  Die  graphische  Statik  der  Axen  und  Wellen."     Published  by 

polytech.  Ver.  in  Zurich,  1863.     [Autograph  copy  of  lectures.] 
Culmann,  K. — "  Der  Werth  der  Constructionen."     Seventh  chap,  of  Graph. 

Statik,  1866.     [Best   and  cheapest  systems  under  given    conditions, 

especially  for  bridges.] 
Reuleaux — "  Graphostatische  Berechnung  verschiedener  Axen,  Kranpfosten, 

Kurbeln,"  in  the  Constructeur,  third  ed.,  1872. 

Scattering  graphostatical  constructions  are  to  be  met  with  in  many  text- 
books upon  construction,  especially  since  the  appearance  of  Culmann's 
work,  a  second  edition  of  which  is  in  course  of  preparation,  and  expected 
soon  to  appear. 


XII. 

GRAPHICAL   DYNAMICS. 

The  scientific  or  practical  value  of  graphical  solutions  once  recognized, 
there  remains  no  reason  for  limiting  them  to  statical  problems  only,  and 
endeavors  in  the  above  direction  are  already  forthcoming.  We  limit  our- 
selves to  a  passing  notice. 

First,  we  have  an  attempt  to  employ  graphical  constructions  in  the 
theory  of  the  overshot  and  breast- wheel  (Qeeberger,  "  Arbeitung  der  Theo- 
rie der  oberschlachtigen  Wasserrader  auf  graphischen  Wege."  Civil  Ing. 
1869,  p.  398,  and  1870,  p.  339).  We  cannot  here  notice  the  value  of  the 
solutions  given,  but  the  very  sparing  applications  of  geometry  hardly  jus- 
tify the  title  of  the  work. 

A  short  article,  which  gives  the  graphical  determination  of  the  force  at 
every  position  of  a  moving  point,  may  also  be  noticed.  (Rapp,  "  Zur 
graphischen  Phoronomie,"  in  Zeitsch.  f.  Math.  u.  Phys.,  1872,  p.  19.) 

The  genuine  foundation  of  a  graphical  dynamics  has  been  first  attempted 
by  Proll  ("  Begriindung  graphischer  Methoden  zur  Losung  dynamische 
Probleme,"  in  Civil  Ingenieur,  1873).  From  the  fact  that  the  effects  of 
forces  in  dynamics  are  measured  by  the  changes  of  velocity  of  any  point  or 
points  of  a  machine  system,  Proll  concluded  that  it  must  be  possible  to 
represent  these  force  effects  by  geometrical  relations,  such  as  kinematic 
geometry  teaches. 

His  investigations,  since  published  in  independent  form  ("Versuch 
einer  graphischen  Dynamic,"  with  10  plates,  1874),  fall  into  thr.ee  parts. 
The  first  part  treats  of  the  action  of  the  "  outer  forces  "  in  machines  whose 


GRAPHICAL   DYNAMICS.  11 

motion  is  in  a  plane,  the  outer  forces  being  also  in  this  plane.  In  the  sec- 
ond part  he  subjects  to  graphical  treatment  the  action  of  outer  forces  upon 
a  free  movable  material  point.  The  third  part,  finally,  considers  the  motion^ 
of  rigid- invariable  systems  acted  upon  by  given  forces. 

In  the  course  of  the  development  extended  use  is  made  of  analytical 
formulae.  The  work  is  but  the  beginning  of  the  future  structure,  but  this 
beginning  will  be  thankfully  received  by  all  those  with  whom  graphical 
methods  have  found  acceptance. 


PART    I. 

GENERAL  PRINCIPLES. 


CHAPTEE  I. 

FORCES   IN   THE   SAME   PLANE COMMON  POINT   OF   APPLICATION. 

1.  Notation,  etc. — In  order  that  a  force  may  be  "  given  "  or 
completely  determined  in  its  relations  to  other  forces,  we  must 
know  not  only  its  intensity,  but  also  its  direction,  and  the  posi- 
tion of  its,  point  of  application.  These  three  being  known,  the 
geometrical  expression  of  our  knowledge  is  very  simple.  We 
have  only  to  assume  a  certain  length  as  the  unit  of  force,  and 
then  any  force  is  at  once  given  by  the  length,  direction,  and 
position  of  a  straight  line.  This  method  of  force  representa- 
tion is  so  obvious,  that  it  is  in  fact  used  in  mechanics,  even 
where  the  treatment  itself  is  essentially  analytical. 

Unless  expressly  stated,  all  the  forces  with  which  we  have  to 
do,  will  be  considered  as  lying  and  acting  in  the  same  plane. 
Graphically  then,  any  force  is  completely  determined  by  a 
straight  line,  the  beginning  of  which  represents  the  point  of 
application,  and  the  length  and  direction  of  which  give  the  in- 
tensity and  direction  of  the  force. 

We  shall  indicate  a  force  in  general  by  the  letter  P,  its  point 
of  application  by  A.  When  we  have  several  forces  we  repre- 
sent the  points  of  application  by  A1?  A2,  A3,  etc.,  and  the  ends 
of  the  corresponding  lines  by  P1?  P2,  P3,  etc.  The  direction  in 
which  a  force  is  supposed  to  act  is  thus  unmistakably  indi- 
cated. 

When,  however,  lines  representing  several  forces  are  laid  off 
one  after  another,  the  beginning  of  each  at  the  end  of  the  pre- 
ceding, it  will  be  sufficient  to  put  0  at  the  beginning  of  the 
first,  and  1,  2,  3,  etc.,  at  the  end  of  each.  No  confusion  can 
arise,  as  each  force  acts  and  reaches  from  the  point  indicated 
by  the  figure  which  is  one  less  than  its  index,  to  the  point  indi- 
cated by  that  index. 

When,  finally,  we  designate  a  force  by  the  two  letters  or  fig- 
ures which  stand  at  the  beginning  and  end,  we  shall  always 
indicate  by  the  order  in  which  the  letters  or  figures  are  written, 


2  FORCES   IN   THE    SAME   PLANE.  [CHAP.  I. 

the  direction  of  action  of  the  force,  first  naming  the  point  of 
application,  and  then  the  end. 

A  force  due  to  the  composition  of  several  forces,  as  Pl5  P2,  P3, 
we  denote  by  Pi.3  or  R^.  Thus  R^  denotes  the  resultant  of 
the  forces  Pl5  P2,  and  P8. 

2.  Parallelogram  of  Force§. — If   two  forces,  P^  and  P2, 
given  in  direction  and  intensify  by  the  lines  OPX  OP2  [Fig.  1, 
PL  1],  have  a  common  point  of  application  O,  the  resultant 
Ri.2  is  found  by  the  well  known  principle  of  the  "  parallelo- 
gram of  forces,"  by  completing  the  parallelogram  as  indicated 
by  the  dotted  lines,  and  drawing  the  diagonal.     OR  then  gives 
the  resultant  of  the  forces  Pl  and  P2.     If  this  resultant  acts  in 
the  direction  from  O  to  R,  as  indicated  by  the  arrow,  it  replaces 
P!  and  P2 ;  that  is,  it  produces  the  same  effect  as  both  forces 
acting  together.     If  it  were  taken  as  acting  in  the  opposite 
direction — i.e.,  from  O  outwards,  away  from  R — it  would  hold 
the  forces  pt  and  P2  in  equilibrium. 

Now,  we  see  at  once  that  it  is  unnecessary  to  complete  the 
parallelogram.  It  is  sufficient  to  draw  from  the  end  of  the 
force  P2  the  line  P2  R  in  the  same  direction  that  Pl  acts  in,  and 
make  it  equal  and  parallel  to  Plf  The  point  R  thus  found  is 
the  end  of  the  resultant  R,  or  is  a  point  upon  the  direction  of 
the  resultant  prolonged  through  O. 

As  to  the  direction  of  action  of  the  resultant — if  we  follow 
round  the  triangle  from  O  to  P2  and  from  P2  to  R  and  R  to  O 
— i.e.,  if  we  follow  round  in  the  direction  of  the  forces — the 
direction  for  the  resultant  from  R  to  O  thus  obtained  is,  as  we 
have  already  seen,  the  direction  necessary  for  equilibrium. 

3.  If,  instead  of  two  forces,  we  have  three  or  more,  as  P1?  P2, 
P3,  P4  [Fig.  2]  we  still  have  the  same  construction.     Thus  com- 
pleting the  parallelogram  for  Px  and  P2  we  find  R^.     Complet- 
ing the  parallelogram  for  R^  and  P8,  we  find  RM,  and  again, 
with  this  and  P4  we  obtain  Rw.     Again,  we  see  it  is  unneces- 
sary to  complete  all  the  parallelograms.     We  have  only  to  draw 
lines  P!  Rj.2,  Ri_2  R^,  Ri_3  RM,  parallel  to  the  forces  P2  P3  and 
P4  respectively,  and  equal  in  length  to  the  intensities  of  these 
forces,  and  then,  no  matter  what  may  be  the  number  of  forces, 
the  line  drawn  from  the  point  of  beginning  to  the  end  of  the 
last  line  laid  off  will  give  the  intensity  and  position  of  the 
resultant.     As  to  direction,  the  same  holds  good  as  before. 

If  the  end  of  the  last  line  laid  off  as  above,  should  coincide 


CHAP.  I.]  COMMON    POINT    OF    APPLICATION.  3 

with  the  point  of  beginning,  there  is,  of  course,  no  resultant, 
and  the  forces  themselves  are  in  equilibrium. 

4.  The  polygon  formed  by  the  successive  laying  off  of  the 
lines  parallel  and  equal  to  the  forces,  we  call  the  "force  poly- 
gon"    Hence  we  have  the  following  principles  established  : 

If  any  number  of  forces  having  a  common  point  of  appli- 
cation and  lying  in  the  same  plane,  a,re  in  equilibrium,  the 
" force  polygon"  is  closed. 

If  the  "force  polygon  "  is  not  closed,  the  forces  themselves 
are  not  in  equilibrium,  and  the  line  necessary  to  close  it  gives 
the  resultant  in  intensity  and  direction. 

This  resultant,  if  considered  as  acting  in  the  direction  ob- 
tained ~by  following  round  the  " force  polygon  "  with  the  forces, 
will  produce  equilibrium — acting  in  the  opposite  direction,  it 
replaces  the  forces. 

The  resultant  thus  found  in  intensity  arid  direction  can  be 
inserted  in  the  force  diagram  at  the  common  point  of  applica- 
tion. 

5.  Thus,  required  the  position,  intensity,  and  direction  of  the 
resultant  of  the  forces  P1?  P2,  P3,  P4,  P5. 

These  forces  are  given  in  position,  direction,  and  intensity 
by  the  force  diagram,  Fig.  3  (a).  The  resultant  of  all  these 
forces  must  have  of  course  the  same  point  of  application  A  as 
the  forces  themselves — it  remains  to  find  then  its  relative  posi- 
tion and  the  direction  of  its  action,  so  that  we  may  properly 
insert  it  in  the  force  diagram. 

We  have  simply  to  draw  the  force  polygon,  Fig.  3,  (b)  by  lay- 
ing off  successively  O  P1?  Pt  P2,  etc.,  equal,  parallel,  and  in  the 
same  direction  as  the  forces  P1?  P2,  etc.,  as  given  by  Fig.  3  (a). 
Then  the  line  P5  O  necessary  to  close  the  force  polygon  gives 
the  intensity  of  the  resultant,  and  in  order  to  replace  P^  it 
must  act  in  the  direction  from  O  to  P5 ;  i.e.,  contrary  to  the 
order  of  the  forces.  If  then  in  Fig.  3  (a)  we  draw  A  Rw  equal 
and  parallel  to  O  P5,  we  have  the  resultant  applied  at  the  com- 
mon point  of  application  A,  and  given  in  position,  intensity 
and  direction. 

Moreover,  it  is  evident  that  any  diagonal  of  the  force  poly- 
gon as  R^  [Fig.  3  (b)]  is  the  resultant  of  P8^,  and  acting  in  the 
direction  from  P4  to  P2,  it  holds  P^  in  equilibrium.  But  it  is 
also  the  resultant  of  Pj,  P2,  P5,  and  R^,  and  acting  in  the  same 
direction  as  before,  it  replaces  these  forces.  The  force  polygon 


FORCES   IN   THE   SAME   PLANE.  [CHAP.  I. 

thus  shows  that  the  force  which  replaces  P1?  P2,  P5,  and  Rw,  at 
the  same  time  holds  P3  and  P4  in  equilibrium,  just  as  it  should 
do. 

If,  on  the  other  hand,  we  had  originally  only  P1?  P2,  R^,  P5, 
and  RLS  forming  a  system  of  forces  in  equilibrium,  we  could 
decompose  Rg^  into  two  components  by  simply  assuming  any 
point  as  P3  [Fig.  3  (£)]  and  drawing  P3  P4,  P3  P2.  Then  follow- 
ing round  this  new  polygon  in.  the  direction  of  the  forces,  or, 
what  amounts  to  the  same  thing,  taking  the  direction  of  the 
components  P3  P4,  opposed  to  the  direction  of  R34  for  equilibri- 
um, we  obtain  the  direction  of  action  of  P3  and  P4  as  shown  by 
the  arrows  in  Fig.  3  (b).  These  forces  inserted  in  Fig.  3  (a),  in 
the  place  of  R3.4  and  in  these  directions,  will  not  disturb  the 
equilibrium. 

Hence,  any  diagonal  in  the  force  polygon,  is  the  resultant 
of  the  forces  on  either  side,  holding  in  equilibrium  those  on 
one  side  and  replacing  those  on  the  other,  according  to  the 
direction  in  which  it  is  conceived  to  act. 

Also,  any  force  or  number  of  forces  may  be  decomposed  into 
two  others  in  any  desired  direction,  ~by  choosing  a  suitable 
point  in  the  plane  of  the  force  polygon  and  drawing  lines 
from  this  point  to  the  beginning  and  end  of  the  force  or  force 
polygon. 

6.  It  matter§  not  in  what  Order  we  lay  off  the  Forees  in 
the  Contraction  of  the  force  Polygon. — Thus,  in  Fig.  1, 
whether  we  draw  from  the  end  of  P2  the  line  P2  R^  equal  and 
parallel  to  P!  or  from  the  end  of  Px  the  line  P!  R^  equal  and 
parallel  to  P2,  in  either  case  we  obtain  the  same  resultant  and 
the  same  direction  for  the  resultant.  But  by  a  similar  change 
of  two  and  two,  we  can  obtain  any  order  we  please.  For  exam- 
ple, we  lay  off  in  Fig.  3  (c)  the  same  forces  in  the  order  P3  P2 
PI,  P5  P4,  and  obtain  precisely  the  same  resultant,  in  the  same 
direction  as  before.  For,  the  resultant  of  P3  and  P2  must  be 
the  same  as  that  of  P2  and  P8  in  the  first  case.  The  resultant 
of  R3.2  and  Pt  must  then  be  the  same  in  both  polygons,  and  so 
on. 

Generally,  then,  no   matter  what  the  order  in  which   the 

forces  are  laid  off,  the  line  necessary  to  close  the  force  polygon 

is  the  resultant  of  the  forces,  and  the  diagonals  of  the  force 

polygon  give  us  the  resultants  of  the  forces  on  either  side. 

By  assuming  a  point  at  pleasure,  and  drawing  lines  from  this 


CHAP.  I.]  COMMON  POINT    OF    APPLICATION.  5 

point  to  the  beginning  and  end  of  any  side  of  the  force  poly- 
gon, and  taking  the  direction  of  these  lines  opposed  to  the 
direction  of  that  side,  we  can  decompose  any  force  in  the  force 
polygon  into  its  components.  Thus  the  force  polygon  gives  ns 
complete  information  as  to  the  action  of  the  forces. 

7.  If  the  Force§  act  in  the  same  straight  I-ine,  the  force 
polygon  of  course  becomes  a  straight  line  also,  and  the  result- 
ant is  the  sum  or  difference  (algebraic  sum)  of  the  forces. 

Thus,  if  we  have  Pl5  P2,  P3,  all  acting  at  the  point  A,  as 
shown  by  the  force  diagram  Fig.  4  (a),  we  form  the  force  poly- 
gon by  laying  off  from  0,  Fig.  4  (&),  the  intensity  of  P1?  from 
the  end  of  this  line  Pl  P2  equal  to  A  P2  and  from  P2,  P2  P3 
equal  to  A  P3.  Then  the  line  necessary  to  close  the  polygon  is 
evidently  0  P3  —  P^  +  P2 — P8.  A  single  force  acting  then  at  A 
in  the  direction  of  and  having  the  intensity  represented  by  the 
line  0  P3  would  replace  P1?  P2,  and  P3.  If  acting  from  P3  to  0, 
it  will  produce  equilibrium. 

If  we  again  choose  an  arbitrary  point  as  C  [we  shall  hereaf- 
ter call  this  point  the  "pole"  of  the  force  polygon],  and  draw 
lines  S0  S3  from  this  pole  to  the  beginning  and  end  of  the  force 
polygon,  we  can  decompose  the  resultant  into  two  forces  in  any 
required  direction.  If  the  resultant  is  supposed  to  act  down, 
then  the  arrows  show  the  direction  in  which  these  components 
must  act  in  order  to  replace  the  resultant.  If  then  at  A  we 
draw  lines  parallel  arid  equal,  we  have  these  components  in  posi- 
tion, direction,  and  applied  at  the  common  point  of  application. 

§.  Practical  Applications. — Simple  and  even  self-evident 
as  all  the  preceding  may  seem,  we  have  already  acquired  all 
the  principles  requisite  for  a  rapid,  accurate,  and  very  elegant 
method  of  finding  by  diagram  the  strains  in  the  various  mem- 
bers of  all  kinds  of  framed  structures,  such  as  roof  trusses, 
bridge  girders,  cranes,  etc.,  no  matter  how  complicated  the 
structure,  or  what  special  assumptions  are  made  as  to  the  load- 
ing, provided  only,  that  all  the  exterior  forces  are  known.  A 
complicated  or  unsymmetrical  arrangement  of  parts  increases 
greatly  the  labor  of  calculation,  but  has  no  effect  upon  the  ease 
or  accuracy  of  the  graphical  method.  The  method  moreover 
checks  its  own  accuracy,  does  not  accumulate  errors,  and  shows 
in  one  view  the  relation  of  the  strains  to  each  other,  and  the 
variations  which  would  be  caused  by  a  change  in  the  manner 
of  load  distribution,  or  in  the  form  of  construction. 


6  FORCES    IN   THE    SAME   P^ANE.  [CHAP.  I. 

As  this  method  is  not  as  well  known  as  it  deserves  to  be,  it 
will  perhaps  be  of  advantage  to  pause  for  a  moment  in  the 
development  of  our  subject,  and  make  this  direct  application  of 
the  principles  already  established. 

BEACED    SEMI-AKCH. 

9.  Stoney,  in  his  "  Theory  of  Strains,"  Yol.  I.,  page  123, 
gives  the  following  example  of  a  "  braced  semi-arch,"  repre- 
sented by  Fig.  5,  PL  1.  The  dimensions  are  as  follows:  pro- 
jecting portion,  40  ft.  long,  10  ft.  deep  at  wall.  Lower  flange, 
circular,  with  a  horizontal  tangent  2  ft.  below  the  extremity  of 
girder.  Radius  of  lower  flange,  104  ft.  Load  uniform  and 
equal  to  one  ton  per  running  foot  supposed  to  be  collected  into 
weights  of  10  tons  at  each  upper  apex,  except  the  end  one, 
which  has  only  5  tons. 

Fig.  5  shows  the  arch  drawn  to  a  scale  of  10  ft.  to  an  inch. 

This  scale  is  too  small  in  this  case  to  ensure  good  results  ;  in 
general  the  larger  the  scale  to  which  the  frame  can  be  drawn, 
the  better;  but  for  the  purpose  of  illustration  it  will  answer 
well  enough.  With  a  large  scale  for  the  frame  diagram,  a 
scale  of  10  tons  to  an  inch  will  in  general  be  found  to  answer 
well.  Fig.  5  (a)  gives  the  strains  in  the  various  members  to  a 
scale  of  10  tons  to  an  inch  and  Fig.  5  (b)  20  tons  to  an  inch ; 
the  first  for  the  load  at  the  extremity  alone,  the  second  for  a 
nniform  load. 

Fig.  5  (a)  is  thus  obtained.  We  first  lay  off  the  weight,  5 
tons,  to  scale,  in  the  direction  in  which  it  acts ;  i.e.,  down- 
wards. Now  this  weight  and  the  strains  in  diagonal  1,  and 
flange  A,  are  in  equilibrium ;  therefore  by  article  (4)  the  force 
polygon  must  close.  Drawing  lines  therefore  from  the  ends  of 
the  line  representing  the  weight  of  5  tons,  parallel  to  these 
pieces  and  prolonging  them  to  their  intersection,  we  obtain 
the  strains  in  A  and  1.  Commencing  with  the  beginning  of 
the  weight  line  and  following  down  around  the  triangle  thus 
formed,  we  find  that  A  acts  from  right  to  left,  as  shown  by  the 
arrow.  A  acts  then  aioay  from  the  apex ;  it  is  therefore  in 
tension.  Diagonal  1  acts  towards  the  apex  and  is  hence  com- 
pressed. 

We  pass  now  to  apex  a,  of  the  frame.  Here  we  have  the 
strains  in  E  and  diagonals  1  and  2,  and  these  three  strains  hold 
each  other  in  equilibrium.  The  strain  in  1  we  have  already, 


CHAP.  I.]  COMMON   POINT   OF   APPLICATION.  7 

and  know  it  to  be  compressive.  We  have  then  simply  to  draw 
lines  from  0  and  b  parallel  to  E  and  2,  and  follow  round  the 
triangle,  to  obtain  the  intensity  and  quality  of  the  strains  in  E 
and  2.  We  must  remember  that  as  1  is  in  compression,  and 
we"  are  now  considering  apex  #,  we  must  follow  round  from  o 
*  to  b  in  Fig.  5  (#),  and  so  round.  We  thus  find  2  acting  away 
from  apex  a  and  therefore  in  tension,  and  E  acting  towards 
this  apex,  and  hence  com/pressed. 

Pass  now  to  apex  c.  We  have  the  strains  in  A  and  2  in 
equilibrium  with  B  and  3.  [No  weights  are  supposed  to  act 
except  the  one  at  the  end.]  But  A  and  2  we  already  have. 
We  draw  3  and  B.  Diagonal  2  has  been  found  to  be  in  ten- 
sion. With  reference  to  apex  c  it  must  therefore  act  away 
fromc;  i.e.,  from  d  to  b  in  the  force  polygon.  This  is  suffi- 
cient to  give  us  the  hint  how  to  follow  round.  We  pass  from 
d  to  b  for  2,  from  b  to  e  for  A,  then  from  e  to  B  and  from  B  to 
d  for  B  and  3.  B  is  therefore  tension  and  3  compression. 
And  so  we  proceed.  For  the  next  apex  g,  we  have  E  and  3  in 
equilibrium  with  F  and  4.  AVe  draw  parallels  to  F  and  4  so 
as  to  close  the  polygon  of  which  we  have  already  two  sides,  E 
and  3,  given,  and  remembering  that  as  3  is  in  compression,  it 
must  therefore  act  towards  g,  we  follow  round  the  completed 
polygon  with  this  to  guide  us,  and  find  4  tension  and  F  com- 
pression.  Thus  we  go  through  the  figure,  and  when  all  is 
ready  we  can  scale  off  the  strains.  The  strains  in  the  lower 
flanges  it  will  be  observed  all  radiate  from  o.  The  upper 
flanges  are  all  measured  off  on  the  horizontal  e  C,  and  the  dia- 
gonals are  the  traverses  between.  We  see  at  once  that  however 
irregular  the  structure,  we  can  always  easily  and  readily  deter- 
mine the  strains  at  any  apex,  provided  no  more  than  two  un- 
known strain^  are  to  be  found.  If  more  than  two  pieces,  the 
strains  in  which  are  unknown,  meet  at  an  apex,  we  can  evi- 
dently form  an  indefinite  number  of  closed  polygons.  The 
problem  is  indeterminate,  and  the  structure  has  unnecessary 
or  superfluous  pieces. 

Fig.  5  (b)  gives  the  strains  for  a  uniform  load,  taken,  for  con- 
venience of  size,  to  a  scale  of  20  tons  to  an  inch.  Here  until 
we  arrived  at  apex  c  of  the  frame  the  strains  are  evidently  the 
same  as  before.  Observe  the  influence  of  the  weight  at  c. 
Here  we  have  the  strains  in  A  and  2  given  in  the  diagram,  in 
equilibrium  with  B,  3  and  the  known  weight  acting  at  c\  viz., 


8  FORCES    IN   THE    SAME   PLANE.  [CHAP.  I. 

10  tons.  We  lay  off  therefore  10  tons  downward  from  e,  Fig. 
5  (£),  and  follow  down  from  e  around  the  polygon.  We  thus 
find  B  tension  and  3  compression.  Then  4  and  F  are  found  as 
before  for  apex  <7,  4  tension  and  F  compression ;  and  then  we 
come  to  the  next  apex  and  the  next  weight.  This  is  laid  off 
downwards  from  the  end  of  the  preceding,  and  then  we  follow 
round,  finding  C  tension  and  5  compression ;  and  so  on. 
1O.  As  another  example,  let  us  take  the 


given  in  Fig.  6,  PI.  2.  This  truss  is  given  by  Stoney,  Yol.  I., 
page  128.  Dimensions :  span,  80  ft. :  rise  of  top  and  bottom 
flanges,  16  and  10  ft.  respectively.  Kadii,  58  and  85  ft.  The 
figure  shows  two  different  kinds  of  bracing.  In  the  left-hand 
part  the  extreme  bay  of  the  lower  flange  is  half  as  long  again 
as  the  others.  The  upper  flange  is  divided  into  4  equal  bays. 
In  the  right-hand  section,  both  flanges  are  divided  into  4  equal 
bays,  and  every  alternate  brace  is  therefore  nearly  radial.  Each 
upper  apex  in  both  cases  is  supposed  to  sustain  a  weight  of 
one  ton. 

The  strains  in  the  various  pieces  are  given  in  Fig.  6  (a). 

We  form  the  force  polygon  by  laying  off  the  weights  from  0 
to  Y  and  then  laying  off  the  reactions  3.5  apiece,  upwards,  we 
come  back  to  0,  and  the  force  polygon  is  closed  as  it  should  be, 
since  the  sum  of  the  reactions  must  be  equal  and  opposite  to 
the  sum  of  the  weights.  Starting  then  with  the  reaction  at  the 
left  support  A,  we  go  through  from  apex  to  apex  in  a  manner 
precisely  similar  to  the  previous  case.  The  operation  is  so 
simple  that  it  is  hardly  necessary  to  detail  it  again,  but  we 
recommend  the  reader  to  go  over  it  with  the  aid  of  Fig.  6  (#), 
lettering  the  figure  as  he  proceeds.  The  dotted  part  gives  the 
strains  for  the  right-hand  half. 

DIAGRAM   FOR   WIND   FORCE. 

11.  It  is  of  considerable  importance  to  investigate  the  influ- 
ence of  a  partial  load,  such  as  that  caused  by  the  wind  blowing 
on  one  side  of  the  roof,  and  this  by  the  aid  of  our  method  we 
can  easily  do. 

From  the  experimental  formulae  of  Hutton,* 

*  Iron  Bridges  and  Roofs.     Unwin.     p.  120. 


CHAP.  I.]  COMMON   POESTT   OF   APPLICATION. 


Pn=P  sin.  i 
Ph=:P  sin.  i  1-84cos-i 
Pv=Pcot.  isin,  i1-84008-1 

where  P  is  the  intensity  of  the  wind  pressure  in  Ibs.  per  sq.  ft. 
upon  a  surface  perpendicular  to  its  direction,  i  is  the  inclination 
of  any  plane  surface  to  this  direction  ;  Pn  is  the  normal  pres- 
sure, Ph  the  horizontal  component  of  this  normal  pressure,  and 
Pv  its  vertical  component. 

That  is,  if  the  wind  blows  horizontally,  Ph  is  the  horizontal 
and  Pv  the  vertical  component  of  the  pressure  on  the  roof.  If 
we  take  P=40  Ibs.,  which  probably  allows  sufficient  margin 
for  the  heaviest  gales,  we  have  the  following  values  of  the  nor- 
mal pressure  and  its  components  for  various  inclinations  of 
roof  surface  : 


Angle  of 
Koof 

Lbs. 

per  square  foot 

of  surface. 

Pa 

Pv 

Ph 

5°  

5.0.  ... 

4.9. 

0.4 

10°  

9.7.... 

9.6. 

1.7 

20°  

18.1  

17.0. 

6.2 

30°  

26.4.... 

22.8. 

13.2 

40°  

33.3.... 

25.5. 

21.4 

50°....... 

38.1  

24.5. 

29.2 

60°  

40.0.... 

20.0. 

34.0 

70°  

41.0.... 

14.0. 

38.5 

80°  

40.4.... 

7.0. 

39.8 

90"  

40.0.. 

0.0. 

40.0 

The  load  at  each  joint  may  be  taken  as  equal  to  the  pressure 
of  the  wind  striking  a  surface  whose  area  is  equal  to  that  por- 
tion of  the  roof  supported  by  one  bay  of  the  rafter,  and  inclined 
at  the  same  angle  as  the  tangent  to  the  rib  at  the  joint.  Thus 
we  can  calculate  P1?  P2,  P3,  P4,  (Fig.  6),  resolve  these  forces  into 
their  horizontal  and  vertical  components,  and  find  the  reactions 
at  the  supports  as  well  as  the  horizontal  force  at  the  left  abut- 
ment, which  in  our  construction  is  supposed  to  be  fixed.  Should 
the  wind  be  supposed  to  blow  from  the  right  side,  the  strains 
would  be  entirely  different,  and  it  would  be  necessary  to  form 
a  second  diagram.  Each  piece  must  be  proportioned  to  resist 
the  strains  arising  in  either  case.  The  forces  P^  and  their 
horizontal  and  vertical  components,  as  also  the  reactions,  being 
known,  we  can  now  form  the  force  polygon. 

Thus  in  Fig.  6  (&),  we  lay  off  the  forces  P4.l5  make  a  c  equal 


10  FOKCES   IN   THE   SAME   PLANE.  [CHAP.  I. 

to  the  vertical  reaction  at  A,  a  b  =  the  sum  of  the  horizontal 
components,  or  the  horizontal  force  at  A,  and  o  1)  the  vertical 
reaction  at  the  right  snppprt.  This  last  line  should  close  the 
force  polygon  and  bring  us  back  to  o. 

Now  starting  at  the  left  support,  we  have  the  vertical  reac- 
tion a  c,  the  horizontal  force  a  b,  and  the  wind  force  Pl5  in 
equilibrium  with  A  and  E.  Closing  the  polygon  by  lines  par- 
allel to  A  and  E,  we  obtain  the  strains  in  these  pieces,  E  ten- 
sion and  A  compression.  At  the  next  apex  we  have  A  and  P2 
in  equilibrium  with  1  and  B.  Completing  the  parallelogram, 
we  find  1  compression  and  B  compression.  At  the  next  apex 
1  and  E  are  in  equilibrium  with*  2  and  F,  and  we  find  F  and  2 
tension  and  so  on.  The  upper  flanges  are  in  compression  and 
start  from  the  ends  of  the  forces  Pl5  P2,  etc.  The  lower  flanges 
radiate  from  b.  If  we  were  to  carry  out  the  construction  for 
the  rest  of  the  frame,  the  upper  flanges  after  D  would  radiate 
from  o. 

A  comparison  of  Fig.  6  (a)  and  (b)  shows  that  whereas  under 
uniform  load  the  strain  in  1  is  tension,  for  wind  force  the  same 
brace  is  in  compression.  In  fact  in  the  first  case  all  the  braces 
are  in  tension,  while  in  the  second  1,  3,  and  5  are  compressed, 
and  3  and  5  quite  severely.  The  strains  in  the  bracing  gener- 
ally are  much  greater  in  the  second  case. 

Were  we  to  consider  the  wind  as  blowing  from  the  other 
side,  or  what  is  the  same  thing,  suppose  the  right  end  fixed  and 
the  left  supported  on  rollers,  then  the  horizontal  reaction  a  b 
will  be  applied  at  the  right  abutment.  In  this  case  the  lower 
flanges  will  radiate  from  a  instead  of  &,  and  the  first  upper 
flange  will  start  from  o.  Supposing  the  first  two  lines  of  this 
new  diagram  drawn,  as  indicated  by  the  dotted  lines,  and  fol- 
lowing round  from  b  to  o,  and  so  round  to  a  and  back  to  5,  it 
may  easily  happen  that  the  last  upper  flange  is  in  tension  and 
the  last  lower  flange  in  compression;  that  is,  a  complete  reversal 
of  the  ordinary  condition  of  strain. 

For  an  excellent  presentation  of  the  above  method,  we  refer 
the  reader  to  Iron  Bridges  and  Roofs,  by  W.  C.  Unwin,  pp. 
128-140.  The  above  method  is  there  referred  to  as  "Prof. 
Clerk  MaxweWs  Method"  and  as  such  is  known  and  used  in 
England.* 

*  Phil.  Mag.,  April,  1864,  and  a  Paper  read  before  the  British  Association  for 
the  Advancement  of  Science,  by  Prof.  Maxwell,  in  1874. 


CHAP.  I.]  COMMON   POINT    OF   APPLICATION.  11 

BRIDGES. 

12.  For  bridges  the  strains  due  to  a  uniform  load  are  of 
course  easily  found.  In  most  cases  a  rolling  load  can  be  man- 
aged also,  without  making  a  separate  diagram  for  each  position 
of  the  load.  Thus,  if  we  diagram  the  strains  for  the  load  at 
the  first  and  last  apex,  the  strains  due  to  intermediate  loads 
will  be  multiples  or  submultiples  of  these.  A  calculation  for  a 
simple  Warren  girder  of  small  span,  and  a  consideration  of  the 
reaction  for  each  position  of  the  load,  will  at  once  illustrate 
what  is  meant.  [Compare  Stoney,  Theory  of  Strains.  Pp. 
99-111,  Yol.  I.] 

Thus  Stoney,  in  his  Theory  of  Strains,  Yol.  I.,  p.  99,  gives 
the  girder  represented  in  Fig.  7,  PL  2,  span  80  ft.,  depth  of 
truss,  5  ft.,  8  equal  panels  in  upper  flange,  7  in  lower. 

For  the  first  weight  of  10  tons,  P1?  the  strains  are  given  by 
Fig.  7  (a]  to  a  scale  of  10  tons  to  an  inch.  We  form  first  the 
force  polygon  by  laying  off  from  o,  10  tons,  to  P^  From  the 
end  of  this  line  we  lay  off  upwards  the  reaction  at  right  abut- 
ment =  •$•  of  10  tons,  or  1.25  tons ;  and  then  the  reaction  at  the 
left  abutment  =  -J  of  10  tons,  back  to  6>,  thus  closing  the  force 
polygon.  [Note. — In  any  structure  which  holds  in  equilibrium 
outer  forces,  the  force  polygon  must  close.  If  it  does  not,  there 
is  no  equilibrium,  and  motion  ensues  (see  Art.  20).]  Com- 
mence now  with  the  reaction  at  a  in  the  frame  diagram,  Fig.  7, 
because  here  we  have  a  known  reaction,  a  o  (force  polygon), 
and  only  two  unknown  strains  to  be  determined.  Drawing 
lines  parallel  to  A  and  1,  we  obtain  the  strains  in  A  and  1. 
Then  pass  on  to  apex  b.  With  the  now  known  strain  in  1,  we 
can  determine  2  and  E. 

Passing  now  to  the  next  apex,  we  have  A  and  2  known,  and 
also  the  weight  Pt.  Join  therefore  Pl  and  E  [Fig.  7  (a)~\  by 
lines  parallel  to  B  and  3.  B  and  3  are  both  in  compression. 
We  find  diagonal  2  also  in  compression,  and  1  in  tension.  That 
is,  ~both  the  diagonals  under  the  weight  are  compressed,  as  evi- 
dently should  be  the  case.  From  4  on  we  have  tension  and 
compression  alternately. 

Fig.  7  (b)  gives  the  strains  due  to  the  last  position  of  the  load 
P7.  The  strains  in  the  diagonals  are  evidently  all  equal,  and 
alternately  tension  and  compression. 

Now  it  is  not  necessary  to  construct  more  than  these  two  dia- 
grams. From  these  two  alone  we  can  determine  the  strains  fur 


12 


FORCES    IN   THE    SAME   PLANE. 


[CHAP.  I. 


any  intermediate  weight.  Thus  scaling  off  the  strains  in  Fig. 
7  (a)  and  (&),  we  can  tabulate  them  under  Px  and  P7,  as  shown 
by  the  table. 


DIAGONALS. 

Pi 

P2 

P3 

P4 

P5 

P6 

PT 

C 

+ 

T 

1        

—12  4 

—10  6 

—  89 

—  71 

—  53 

—  35 

1  8 

49  6 

2 

+  12  4 

+  10  6 

+•89 

' 

+  71 

+  53 

+  35 

+  18 

+  49  6 

3  

+  1.8 

—10.6 

—  8.9 

—  71 

—  53 

—  3.5 

—  18 

+  18 

—37  2 

4  

—  18 

+  10.6 

+  89 

+  7.1 

+  53 

+  35 

+  18 

+  37  2 

—  18 

5     .... 

+  18 

+  35 

—  89 

—  7  1 

—  5.3 

—  35 

—  1.8 

+  53 

26  6 

6 

—  18 

—  35 

+  89 

+  7.1 

+  53 

+  35 

+  18 

+  26  6 

53 

7 

+  18 

+  35 

+  53 

—  71 

—  53 

—  35 

—  18 

+  10  6 

—17  7 

8  

-  1.8 

—  35 

—  53 

+  7.1 

+  53 

+  35 

+  1.8 

+  17.7 

—10  6 

Now  the  reaction  at  the  left  abutment  due  to  P6  is  twice 
as  great  as  that  due  to  P7.  Hence  the  values  in  the  column 
for  P6  will  be  twice  as  great ;  in  the  column  for  P5  three  times 
as  great,  and  so  on.  For  similar  reasons  the  strain  in  5  for 
P2  will  be  twice  that  for  P^  In  column  P2,  then,  from  5 
down  we  multiply  the  strains  in  P1  by  2.  In  P8  from  7  down 
by  3.  Thus  we  fill  out  the  table  of  strains  completely,  and  find 
the  maximum  tension  and  compression.  A  similar  procedure 
will  give  the  flanges.* 

APPLICATION   TO   AN   AECH. 

13.  For  a  "  braced  arch  "  (Stoney,  p.  136)  as  represented  in 
Fig.  5  (c)  PL  2,  the  strains  in  every  piece  due  to  any  load  are 
in  similar  manner  easily  found  by  first  finding  the  components 
of  the  load  acting  at  the  abutments,  and  then  proceeding  as 
above.  Thus  for  a  load  P2,  the  left  half  of  the  arch  is  in  equi- 
librium with  the  forces  acting  upon  it ;  viz.,  a  horizontal  and  a 
downward  force  at  a,  and  a  horizontal  and  an  upward  force  at 
A.  The  resultant  of  the  forces  at  a  must  then  pass  through 

*  The  reader  not  familiar  with  the  above  method  of  tabulation  will  find  it 
further  illustrated  in  Art.  7  of  the  Appendix.  He  cannot  do  better  than  to 
refer  to  it  here  and  now. 


CHAP.  I.]  COMMON    POINT    OF   APPLICATION.  13 

a  and  A,  and  be  equal  and  opposite  to  the  resultant  at  A.  The 
resultant  at  the  right  abutment  must  pass  through  that  abutment, 
and  also  through  the  intersection  of  P2  with  A  a.  So  for  any 
other  force,  as  P6,  we  have  simply  to  draw  B  a  to  intersection 
with  P6,  and  then  P6  A.  We  can  now  decompose  P6  or  P2  along 
the  resultants  through  the  abutments  thus  found.  Thus  resolv- 
ing P2  along  A  a  and  P2  B,  Fig.  5  (e\  we  find  the  force  acting  at 
apex  a.  This  force  resolved  into  A  and  1  gives  the  strains  on 
these  pieces  both  compressive.  Passing  then  to  the  next  apex, 
we  obtain  the  strains  in  2  and  E.  Then  to  the  next,  and  we 
get  3  and  B,  compression  and  tension  respectively,  and  so  on, 
as  shown  by  diagram,  Fig.  5  (<?),  which,  it  will  be  seen  at  once, 
is  similar  to  Fig.  5  (a),  already  obtained  for  the  "  semi-arch," 
except  that  the  strain  in  A  is  less  than  for  the  semi-arch  and 
compressive,  while  B  C  and  D  are  in  tension.  The  reason  is 
obvious.  At  a  [Fig.  5  (c)]  the  resultant  lies  between  A  and  1, 
and  therefore  causes  compression  in  both,  while  it  passes  out- 
side of  the  arch  entirely,  to  the  right  of  the  apex  for  diagonals 
3  and  4,  and  hence  causes  tension  in  B  C  and  D.  Fig.  5  (d) 
gives  the  strains  due  to  P6.  Here  the  resultant  or  reaction  at 
A  is  first  found  and  resolved  into  9  and  H,  and  then  we  go 
through  the  frame  as  before.  We  see  that  4  and  5  under  the 
load  are  both  compressed,  that  E  and  F  are  in  tension  and  G- 
and  H,  as  also  the  entire  upper  chord,  in  compression.  The 
work  checks  from  the  fact  that  the  line  closing  the  polygon 
formed  by  E  and  2  should  be  exactly  parallel  to  and  give  the 
strain  in  diagonal  1,  or  A  and  1  should  be  in  equilibrium  with 
the  resultant  through  a  [see  Fig.  5  (d)~\. 

In  every  case  of  the  kind  we  first,  then,  have  to  draw  the 
frame  diagram.  Then  lay  off  \\\Q  force  polygon  which  should 
close.  Finally  we  construct  the  strain  diagram.  The  frame 
diagram  should  be  taken  to  as  large  a  scale  as  possible  consist- 
ent with  reasonable  size,  and  the  scale  for  the  force  and  strain 
diagrams  as  small  as  possible,  consistent  with  scaling  off  the 
strains  to  the  requisite  degree  of  accuracy.  A  small  frame 
diagram  does  not  give  with  the  proper  accuracy  the  relative 
positions  and  inclinations  of  the  various  pieces,  so  as  to  ensure 
the  proper  direction  for  the  lines  of  the  strain  diagram.  A 
slight  deviation  from  parallelism  causes  sometimes  considerable 
variation.  Nevertheless  with  practice,  care,  and  proper  instru- 
ments the  accuracy  of  the  method  is  surprising  ;  even  in  com- 


14  FORCES   IN   THE    SAME   PLANE.  [CHAP.  I. 

plicated  structures,  the  variation  resulting  from  performing 
the  operation  twice  being  inappreciable.  Every  symmetrical 
frame  gives  also  a  symmetrical  strain  diagram,  and  the  accu- 
racy of  the  work  is  tested  at  every  point  by  this  double  sym- 
metry, and  finally  by  the  end  or  last  point  of  the  second  half, 
exactly  coinciding  with  the  last  point  of  the  first  half.  Thus 
in  Fig.  6  (a),  if  we  had  but  one  system  of  triangulation  carried 
through  the  frame,  the  strain  diagram  for  the  right  half  would 
be  precisely  similar  and  symmetrical  to  that  already  found  for 
the  first,  and  the  end  of  the  last  line  would  fall,  or  should  fall, 
precisely  upon  the  point  b  of  the  first.  If  it  does  not,  and  the 
error  is  too  great  to  be  disregarded,  then  by  checking  corre- 
sponding points  in  each  half,  we  can  find  the  point  where  the 
error  was  committed.  In  any  case  errors  do  not  accumulate. 
Thus,  armed  with  straight  edge,  scale,  triangle,  and  dividers, 
we  can  attack  and  solve  the  most  intricate  problems,  without 
calculation  or  tables,  with  ease,  accuracy,  and  great  saving  of 
time. 

METHOD   OF   SECTIONS. 

14.  The  results  obtained  by  the  above  method  are  best 
checked  in  general  by  Hitter's  u  method  of  sections,"  or  the 
use  of  moments.*  This  consists  in  supposing  the  structure 
divided  by  a  section  cutting  only  three  pieces.  We  can  then 
take  the  intersection  of  two  of  these  pieces  as  a  centre  of  mo- 
ments, and  the  sum  (algebraic)  of  the  moments  of  all  the 
exterior  forces,  such  as  reaction,  loads,  etc.,  upon  one  of  the 
portions  into  which  the  structure  is  divided  by  the  section,  with 
reference  to  this  centre  of  moments,  must  be  balanced  by  the 
moment  of  the  strain  in  the  third  piece,  with  reference  to  this 
same  point.  Thus  in  Fig.  6,  PI.  2,  required  the  strain  in  D. 
Take  a  section  through  D,  7  and  H  (right  half  of  Fig.),  and  let 
a  be  the  centre  of  moments.  The  moments  of  the  strains  in  7 
and  H  are  then,  of  course,  zero,  since  these  pieces  pass  through  a. 
The  moment  of  the  strain  in  D  with  reference  to  a  must  then 
be  balanced  by  the  sum  of  the  moments  of  all  the  outer  forces 
acting  upon  the  portion  to  the  left  (or  right)  of  the  section. 

Thus,  strain  in  D  multiplied  by  its  lever  arm  with  respect  to 
#,  is  equal  to  moment  of  reaction  at  A,  minus  sum  of  the  mo- 
ments of  loads  between  A  and  #,  all  with  reference  to  a.  If 

*  Dock-  und  Brucken- Constructional.    Ritter.    Hannover,  1873. 


CHAP.  I.]  COMMON    POINT    OF   APPLICATION.  15 

we  take  the  direction  of  rotation  of  the  forces  on  the  left  of  the 
section  when  in  the  direction  of  the  hands  of  a  watch  as  posi- 
tive^ and  find  the  moment  of  strain  in  D  negative,  it  shows  i 
negative  rotation  about  a,  and  the  strain  in  D  to  resist  this  rota- 
tion must  act  away  from  &,  or  be  tensile.  If  the  resultant 
rotation  of  the  outer  forces  is  on  the  other  hand  positive,  the 
strain  in  D  must  act  toward  b,  and  D  is  therefore  compressed. 

This  method  of  calculation,  it  will  be  observed,  is  both  sim- 
ple and  general.  It  can  be  applied  to  any  structure,  when  the 
outer  forces  are  completely  known,  and  only  three  pieces  are 
cut  by  the  ideal  section. 

15.  It  is  unnecessary  to  give  here  further  applications  of  our 
graphical  method.  The  reader  can  easily  apply  it  for  himself 
to  the  "  bowstring  girder,"  bent  crane,  etc.,  and  satisfy  himself 
as  to  its  accuracy,  and  the  ease  with  which  the  desired  results 
are  obtained. 

Enough  has  been  said  to  indicate  the  many  important  appli- 
cations which  even  at  the  very  commencement  of  our  develop- 
ment of  the  graphical  method  we  are  enabled  to  make,  and 
here  we  shall  close  our  discussion  of  forces  lying  in  the  same 
plane  and  having  a  common  point  of  application.  As  we  pass 
on  to  forces  having  different  points  of  application,  we  shall 
have  occasion  to  develop  new  principles  and  relations  not  less 
fruitful  and  useful  in  their  practical  results.* 


*  We  refer  the  reader  here  to  the  Appendix  to  this  chapter  for  further 
illustrations  of  the  application  of  the  above  principles,  as  well  as  for  informa- 
tion upon  several  points  of  considerable  practical  importance.  We  would  also 
remind  him  here  once  for  all,  that  the  Appendix  to  this  work  was  NOT  in- 
tended to  be  disregarded,  but  has  been  thought  desirable  in  order  to  avoid 
encumbermg  the  general  principles  with  too  much  of  detail  in  the  text.  We 
earnestly  request  him  to  neglect  no  reference  to  it  which  may  be  made  in  the 
text. 

He  will  do  well  in  the  present  case,  after  first  making  himself  familiar  with 
the  above  points,  to  solve  for  himself  with  scale  and  dividers  a  number  of 
similar  problems,  checking  his  results  always  by  the  method  of  moments. 
He  will  thus  in  a  very  short  time  master  the  method,  and  be  able  to  solve 
readily  and  accurately  every  problem  of  usual  occurrence  in  practice. 
Though  the  method  is  very  simple,  actual  practice  with  the  drawing  board  is 
here  indispensable. 


16  FORCES   IN   THE    SAME   PLANE.  [CHAP.  II. 


CHAPTER    II. 


FORCES   IN   THE   SAME   PLANE DIFFERENT   POINTS    OF   APPLICATION. 

16.  Re§ultant  of  Two  Forces  in  a  Plane— Different 
Points  of  Application. — Heretofore  we  have  considered 
forces  having  a  common  point  of  application,  and  have  seen 
that  in  any  case  the  direction  and  intensity  of  the  resultant  is 
easily  found  by  closing  the  force  polygon. 

But  suppose  we  have  two  forces  Pl  P2  having  different 
points  of  application  At  A2 ;  required  the  position  and  direc- 
tion of  the  resultant  [PI.  3,  Fig.  8]. 

Any  force  acting  in  a  plane  may  be  considered  as  acting  at 
any  point  in  its  line  of  direction. 

P!  and  P2  may  then  be  supposed  to  act  at  their  common 
point  of  intersection  #,  and  through  this  point  the  resultant 
should  pass.  The  case  reduces  therefore  to  a  common  point 
of  application.  The  resultant  is  given  in  intensity  and  direc- 
tion as  before  by  the  force  polygon  (&),  and  its  position  is  deter- 
mined by  the  point  of  intersection  a.  At  this  point,  or  at  any 
point  in  the  line  through  a,  parallel  to  0  2,  the  resultant  may 
be  supposed  to  act. 

But  the  direction  of  the  forces  may  not  intersect  within 
reasonable  limits,  or  the  forces  may  be  supposed  parallel  to 
each  other,  so  that  they  may  not  intersect  at  all.  In  any  case 
the  force  polygon  will  still  give  the  intensity  and  direction  of 
action  of  the  resultant,  but  its  position  in  the  plane  of  the 
forces  remains  yet  to  be  determined.  Now  we  have  se.en  [Art. 
5]  that  we  can  decompose  a  force  into  two  components  in  any 
desired  directions,  by  choosing  a  "pole  "  and  drawing  lines  to 
the  beginning  and  end  of  the  force  in  the  force  polygon.  Let 
us  choose  then  a  pole  C  [Fig.  8  (5)]  and  decompose  the  result- 
ant thus  into  two  forces  given  in  intensity  by  the  lines  0  C 
and  2  C.  The  forces  Px  P2  being  supposed  to  act  at  the 
points  A!  A2  in  the  common  plane,  at  what  point  in  the  plane 
and  in  what  direction  must  the  resultant  0  2  be  applied  to  keep 


CHAP.  II.]  DIFFERENT   POINTS    OF    APPLICATION.  17 

this  plane  and  hold  the  forces  in  equilibrium?  The  direction 
of  action  of  the  resultant  is  given  at  once  from  the  force  poly- 
gon  [Art.  5  (&)].  It  must  act  in  a  direction  from  2  to  0,  and 
must  be  equal  to  2  0,  taken  to  the  scale  of  force.  Now  at  any 
point  in  the  line  of  direction  of  P1?  as  for  instance  1,  let  us 
suppose  the  component  given  by  C  0  to  act.  What  is  then  the 
resultant  of  P!  and  CO?  A  glance  at  the  force  polygon  gives 
us  1C,  because  this  line  closes  the  polygon  made  by  C  0,  0  1 
and  1C.  At  I  then,  the  three  forces  S0  (parallel  and  equal  to 
C  0)  St  (parallel  and  equal  to  1  C)  and  P!  are  in  equilibrium, 
and  there  is  no  tendency  of  the  point  1  to  move.  But  1  C  or 

51  may  be  considered  as  acting  in  the  plane  at  any  point  in  its 
line  of  direction ;  therefore  at  2  its  intersection  with  P2  pro- 
longed.    Suppose  at  2,  S2  or  2  C  to  act.     "We  see  at  once  from 
the  force  polygon  that  2  C,  C  1  and  P2  are  in  equilibrium. 
There  is  therefore  no  tendency  of  the  point  2  to  move,  and  the 
two  forces  Pl  P2  are  then  in  equilibrium  with  C  0,  1  C,  C  1 
and  2  C.     But  since  the  resultant  of  C  0  and  2  C  or  of  S0  and 

52  is  also  the  resultant  of  the  forces,  and  since  it  must  there- 
fore act  through  the  point  of  intersection  of  S0  and  S2 :  we 
have  only  to  prolo ng  these  lines  to  intersection  b.      Through 
this  point  the  resultants  R^2  must  pass  and  acting  downwards 
(from  0  to  2)  as  indicated  in  the  Fig.,  it  replaces  Pl  P2.     Act- 
ing upwards  it  would  hold  them  in  equilibrium.     We   thus 
easily  find  the  point  2  in  the  plane  at  which  2  C  or  S2  must 
be  applied,  when  C  0  or  S0  acts  at  1,  and  S0  ?2  are  thus  found 
in  proper  relative  position.     The  position,  intensity,  and  direc- 
tion of  the  resultant  are  thus  completely  determined. 

Had  we  taken  ai*y  other  point  than  1,  as  the  point  of  applica- 
tion of  C  0,  we  should  have  found  a  different  corresponding 
point  for  application  of  2  C,  but  in  any  case  the  prolongations 
of  2  C  and  C  0  would  intersect  upon  the  line  a  b,  prolonged  if 
necessary.  The  same  holds  true  for  any  position  of  the  "pole  " 
C.  This  construction  is  evidently  general  whatever  the  posi- 
tion or  whatever  the  number  of  the  forces.  We  may  thus 
obtain  any  number  of  points  along  the  line  a  b ;  that  is,  the 
resultant  also,  may  act  at  any  point  in  its  line  of  direction. 

[NOTE. — That  b  is  a  point  in  the  resultant  of  Pt  andP2  can 
be  proved  in  a  method  purely  geometrical.  In  the  two  "  com- 
plete quadrilaterals  "  0  1  2  C  and  1  b  2  #,  the  Jive  pairs  of 
corresponding  sides  0  1  and  a  1,  1  2  and  a  2,  2  C  and  b  2,  C  0 


18  FORCES    IN    THE    SAME   PLANE.  [CHAP.  II. 

and  b  1,  C  1  and  1  2,  are  parallel  each  to  each,  therefore  the 
sixth  pair  0  2  and  a  b  must  also  be  parallel  /  b  is  therefore  a 
point  of  the  resultant  passing  through  a,  parallel  to  0  2.] 

17.  The  above  Construction  holds  good  equally  well  for 
Parallel  Forces. — By  means  of  it  we  find  in  PL  3,  Fig.  9  (a) 
and  (b)  and  Fig.  10  (a)  and  (&),  the  resultant  of  a  pair  of  paral- 
lel forces,  in  the  first  case,  both  acting  in  the  same  direction ; 
in  the  second,  in  opposite  directions. 

In  both  cases  we  have  simply  to  choose  a  pole  C,  and  draw 
S0  Sj  and  S2.  Then  taking  any  point  c  in  the  line  of  direc- 
tion of  Pl5  as  a  point  of  application  for,  S0,  draw  through  this 
point  Sl5  thus  finding  d,  the  point  of  application  for  S2.  S0 
and  S2  prolonged,  intersect  upon  the  resultant,  whose  intensity, 
direction,  and  position  thus  become  fully  known. 

18.  Property  of  the  Point  b. — It  is  plain  that  thus  a  point 
of  intersection  J,  through  which  the  resultant  must  pass,  can 
always  be  found,  provided  S0  and  S2  do  not  fall  together  in 
the  force  polygon,  or  intersect  without  the  limits  of  the  draw- 
ing.    By  properly  choosing  the  position  of  the  pole  C,  this  can 
always  be  avoided  if  the  points  2  and  0  in  the  force  polygon  do 
not  themselves  coincide,  i.e.,  if  the  force  polygon  does  not  close. 

The  point  &,  Figs.  8,  9,  and  10,  which  by  reason  of  the  arbi- 
trary position  of  the  pole  may  lie  anywhere  upon  the  resultant, 
has  a  remarkable  property.  If  we  draw  a  line  m  n  through 
this  point  parallel  to  S^  and  let  fall  from  it  perpendiculars  pt 
and  p2  upon  Pl  and  P2,  then  in  all  three  cases,  and  therefore 
generally,  the  triangle  c  m  b  is  similar  to  0  C  1,  and  d  b  n  is 
similar  to  1  C  2.  Hence  we  have  the  proportions — 

0  1  :  1  C  I ".  c  m  :  m  b,  and 

1  C  :  12  ; :  n  b  :  n  d. 

From  these  proportions  we  find 

0  ~L  :  I  2  \\  c  m  x  nb  :  mb  x  n  d. 

Now  the  triangles  c  m  b  and  d  n  b  have  the  same  height 
above  the  base  m  n ;  the  bases  m  b  and  b  n  are  therefore  pro- 
portional to  their  areas.     But  their  areas  are  equal  to  half  their 
sides  cm  and  n  d  multiplied  byj?i  and^>2  respectively.     Hence 
we  have  from  the  above  proportion,  since  c  m  =  n  d, 
0  1  :  1  2  ; :  n  d  x  pz :  n  d  x  pt  or 
01:12  ::pz:p,. 
That  is,  the  perpendiculars  let  fall  from  any  point  of  th/>, 


CHAP.  II.]  DIFFERENT   POINTS    OF   APPLICATION.  19 

resultant  upon  the  components,  are  to  each  other  inversely  as 
the  components.  Regarding  any  point  of  the  resultant  as  a 
centre  of  moments,  the  moments  of  the  forces  then  are  equal, 
and  of  course  the  forces  themselves  are  inversely  as  their  lever 
arms. 

19.  Equilibrium  Polygon. — If  we  consider  the  forces  Pt 
P2,  Figs.  8,  9,  and  10,  held  in  equilibrium  by  their  components 
C  0,  1  C,  and  2  C,  C  1,  which  act  parallel  to  the  lines  S0  Sx 
and  S2 ;  then  regarding  the  line  St  or  c  d  as  part  of  the  mate- 
rial plane  in  which  the  forces  act,  C  1  and  1  C  balance  one 
another,  and  cause  either  tension  or  compression  in  c  d.  Sup- 
pose the  resultant  R  is  to  act  so  as  to  cause  equilibrium,  or 
prevent  the  motion  of  the  plane  due  to  Pt  and  P2.  Then  R 
must  act  upwards  in  Figs.  8  and  9,  and  downwards  from  2  to 
0  in  Fig.  10.  In  Figs.  8  and  9  then,  S0  and  S2  act  away  from 
c  and  d  (Art.  4),  and  in  Fig.  10  towards  c  and  d.  Following 
round  the  force  polygon,  we  find  in  the  first  two  cases  c  d  in 
tension,  in  the  last  c  d  in  compression. 

In  the  first  two  cases,  the  points  of  application  c  and  d  of  S0 
P!  and  S2  P2  if*  connected  by  a  string  stretched  between  c 
and  d  will  be  perfectly  fixed  and  motionless  ;  while  in  the  lat- 
ter case,  the  string  must  be  replaced  by  a  strut.  In  case  of 
three  or  more  forces  the  polygon  or  broken  line  which  we  thus 
obtain,  by  choosing  a  pole,  drawing  lines  to  the  beginning  and 
end  of  the  forces  in  the  force  polygon,  and  then  parallels  to 
these  lines  intersecting  the  lines  of  direction  of  the  forces  in  the 
force  diagram,  we  call  the  "  string "  or  "funicular  polygon" 
or  the  " strut  polygon"  according  as  the  forces  act  to  cause 
tension  or  compression  along  these  lines.  We  can  apply  to 
both  cases  the  general  designation  of  polygon  of  equilibrium  or 
"  equilibrium  polygon"  The  perpendicular  let  fall  from  the 
pole  C  upon  the  direction  of  the  resultant  in  the  force  polygon, 
we  call  the  "pole  distance  "and  shall  always  designate  it  by 
H.  The  straight  line  joining  the  points  c  and  d,  or  the  begin- 
ning and  end  of  the  equilibrium  polygon,  we  call  the  "strut" 
or  "  tie  line  "  or  generally  the  "  closing  line  "  and  designate  it 
by  L.  The  convenience  and  '  application  of  these  terms  and 
conceptions  will  soon  appear.  In  the  present  case  of  only  two 
forces,  the  equilibrium  polygon  becomes  a  straight  line  and 
coincides  with  L,  or  c  d. 

[XoTE. — We  repeat  that  in  order  to  determine  the  quality  of 


20  FORCES   IN   THE   SAME   PLANE.  [CHAP.  II. 

the  strain  in  c  d,  we  have  only  to  follow  round  the  force  poly- 
gon in  the  direction  of  the  forces,  and  then  refer  to  the  force 
diagram.  Thus  Fig.  9,  at  c,  Px  S0  and  Sx  act,  and  are  in  equi- 
librium. The  corresponding  closed  figure  is  given  in  the  force 
polygon  (a).  S0  acts  away  from  c,  Px  acts  downwards  from  0  1. 
Continuing  this  direction  we  find  Sx  acting  from  1  towards  C. 
Reversing  this  direction  (Art.  4),  we  find  that  the  resultant 
which  replaces  S0  and  Pt  acts  from  C  to  1.  Referring  now  to 
the  force  diagram  (&),  and  transferring  this  direction  to  the 
point  c,  we  find  this  resultant  acts  to  pull  c  away  from  d  or 
contrary  to  the  direction  of  the  force  1  C  which  replaces  S2  and 
P2.  The  strain  in  c  d  is  therefore  tension. 

A  much  better  way  of  arriving  at  the  same  result  is  to  con- 
sider the  triangle  c  b  d  as  a  jointed  frame  which  holds  in  equi- 
librium the  forces  P!  P2  and  R^.  Then  the  strains  in  any  two 
pieces  c  d,  c  ft,  meeting  at  a  point,  are  in  equilibrium  with  the 
force  or  forces  acting  at  that  point. 

We  have  then  the  force  Pj  acting  at  apex  c,  decomposed  into 
strains  along  c  b  and  c  d  (Art.  5).  represented  by  C  0  and  1  C  in 
the  force  polygon.  All  three  are  in  equilibrium.  Pl  acts 
down.  Follow  down  then  from  0  to  1  from  1  to  C  and  C  to  0. 
Refer  back  now  to  apex  c  of  the  frame  and  transfer  these 
directions.  The  strain  in  c  d  acts  away  from  the  apex  c  and  is 
therefore  in  tension,  while  the  piece  c  b  would  be  in  compres- 
sion, since  the  direction  of  C  0  is  towards  apex  c. 

See  also  "  practical  applications "  of  the  preceding  chapter 
for  illustrations  of  this.  In  the  same  way  follow  round  0  1  C 
Fig.  10  (a)  and  refer  to  (b)  and  S0  is  in  compression^ 

2O.  Ca§e  of  a  Couple. — In  Article  18  we  remarked  that  the 
pole  can  always  be  chosen  in  such  a  position  as  to  give  S0  and 
S2  intersecting  within  desired  limits,  provided  that  S0  and  83  or 
the  point  0  and  2  do  not  coincide.  This  case  however  actually 
happens,  with  a  pair  of  equal  and  opposite  forces — that  is,  with 
a  couple. 

Thus  in  Fig.  11,  PI.  3,  we  have  two  equal  and  opposite  force's 

PI,  P2- 

The  force  polygon  closes :  therefore  the  resultant  is  zero. 
S0  and  83  are  parallel,  hence  their  point  of  intersection  in  the 
equilibrium  polygon  is  infinitely  distant.  By  changing  the 
position  of  the  pole,  we  see  that  S0  and  S2  may  take  any  posi- 
tions in  the  plane. 


CHAP.   II.]  DIFFERENT   POINTS    OF   APPLICATION.  21 

Two  forces  therefore  which  form  a  couple  cannot  be  replaced 
by  a  single  force.  Their  resultant  is  an  indefinitely  small  force 
situated  in  any  position  in  the  plane  of  the  forces,  at  an  infinite 
distance. 

Conditions  of  Equilibrium. — If  then,  similarly  to  Art.  4, 
any  number  of  forces  lying  in  the  same  plane  and  having  differ- 
ent points  of  application,  are  in  equilibrium,  the  force  polygon 
always  closes. 

For  this  reason,  as  already  repeatedly  seen  in  the  practical 
applications  of  our  last  chapter,  the  force  polygon  form'ed  by 
the  exterior  forces  must  always  close. 

But  inversely,  if  the  force  polygon  closes,  it  does  not  follow 
that  the  forces  are  in  equilibrium — a  couple  may  result. 

To  determine  whether  this  is  the  case  inspect  the  "  equilibri- 
um polygon."  If  this  also  closes  [i.e.,  if  S0  and  Sn  intersect] 
the  forces  are  in  equilibrium.  If  this  does  not  close  [i.e.,  if  S0 
and  Sn  are  parallel]  there  is  no  single  resultant,  but  the 
forces  can  be  replaced  by  a  couple,  and  this  couple,  as  we  have 
seen,  may  have  any  position  in  the  plane. 

21.  Thus  if  we  suppose  in  Fig.  11,  PI.  3,  Pt  and  P2  decom- 
posed into  their  components  S0,  S1?  and  S1?  S2,  the  compressive 
strains  in  Sl  at  c  and  d  are  equal  and  opposite  [see  (a)].  We 
have  then  S0  and  S2  remaining,  which  again  form  a  couple 
which  must  have  the  same  action  as  the  first. 

Hence  we  see  that  one  couple  can  be  replaced  by  another  with- 
out changing  the  action  of  the  forces. 

It  is  easy  to  determine  a  simple  relation  between  any  two 
couples. 

If  from  c  we  lay  off  c  a  equal  to  o  1,  and  c  o  equal  to  Co,  we 
have  o  a  parallel  to  C  1  or  S1?  and  therefore  to  c  d.  Join  a  d  and 
o  d.  The  triangles  c  d  a  and  c  d  o  having  a  common  base  c  d 
and  their  vertices  o  and  a  in  a  line  parallel  to  c  d,  are  equal  in 
area.  The  side  c  a  of  one  is  known,  and  the  opposite  apex  lies 
in  the  line  of  the  force  P2.  Its  area  is  then  ca.=  Pl  multiplied 
by  half  of  the  perpendicular  distance  of  Pt  from  P2,  and  is 
therefore  completely  determined.  So  also  for  the  other  trian- 
gle, one  side  of  which  o  c  is  one  force  of  the  new  couple,  and 
the  opposite  apex  of  which  lies  in  the  other  force  S2. 

Hence — a  couple  can  be  turned  at  will  in  its  plane  of  action, 
and  the  intensity  and  direction  of  its  forces  can  be  changed  at 
will  if  the  area  of  the  triangle  the  base  of  which  is  one  of  the 


22  FOSCES    IN   THE    SAME    PLANE.  [CHAP.  II. 

new  forces,  and  whose  opposite  apex  lies  in  the  other  force,  is 
constant;  or  when  the  product  of  the  intensity  of  the  forces 
into  their  perpendicular  distance  remains  the  same.  The  di- 
rection *of  rotation,  of  course,  must  also  remain  the  same. 

We  shall  see  further  on  the  significance  of  this  area,  or  of 
this  product — so  much  is  clear,  that  a  couple  (or  infinitely  small, 
infinitely  distant  force)  is  completely  determined  in  its  plane 
when  the  direction  of  rotation  is  given,  and  the  area  of  the  tri- 
angle or  value  of  the  product  to  which  it  is  proportional,  is 
known.  The  couple  itself  can  be  replaced  by  any  two  parallel 
equal  and  opposite  forces  whatever,  if  only  the  triangle  having 
one  force  as  base,  and  the  opposite  apex  in  the  other,  has  a  given 
constant  area.* 

22.  Force  and  Equilibrium  Polygons  for  any  Number 
of  Force§  in  a  Plane. 

In  PL  3,  Fig.  12  (b)  we  have  the  forces  Pw  acting  in  various 
directions  and  at  different  points  of  application.  P2  and  P3 
form  a  couple ;  that  is,  are  equal,  parallel,  and  opposite  in  di- 
rection. Required  the  position,  intensity  and  direction  of  action 
of  the  resultant. 

First,  form  the  force  polygon,  Fig.  12  (a\  by  laying  off  the 
forces  to  scale  one  after  the  other  in  proper  direction.  Thus 
we  have  0  1,  1  2,  2  3,  3  4,  4  5  in  Fig.  12  (a)  parallel  respec- 
tively to  P!  P2  P3,  etc.,  in  Fig.  12  (b).  The  line  necessary  to 
close  the  polygon,  0  5,  is  the  resultant  in  intensity  and  direc- 
tion. In  intensity  because  the  length  of  0  5  taken  to  the  scale 
of  force,  gives  the  intensity  of  the  resultant ;  in  direction 
because  acting  from  5  to  0  it  produces  equilibrium,  while  act- 
ing in  the  opposite  direction,  from  0  to  5,  it  replaces  the  forces. 

"We  have,  therefore,  only  to  find  the  position  of  the  resultant 
in  the  plane  of  the  given  forces  in  Fig.  12  (b).  Hence : 

Second,  choose  anywhere  a  "pole  "  as  C,  and  draw  the  lines 
or  rays,  or  "  strings  "  S0  St  S2  S3  S4,  etc.  S0  and  S5  are  evi- 
dently components  of  the  resultant,  since  they  form  with  it  a 
closed  figure  in  the  force  polygon. 

Third,  form  the  equilibrium  polygon  a  b  cde  o',  Fig.  12  (J), 
as  follows : 

Draw  a  line  parallel  to  S0  intersecting  T?±  (produced  if  neces- 
sary) at  any  point  as  a.  From  this  point  draw  a  line  parallel 

*  Eltmente  der  Graphischen  Statik.  Bauschinger.  Munchen.  1871.  Pp. 
11,  12. 


CHAP.  II.]  DIFFERENT   POINTS    OF    APPLICATION.  23 

to  Si  to  intersection  with  P2  (also  produced  if  necessary)  at  b. 
From  b  parallel  to  S2  to  c,  then  parallel  to  S3  to  d,  and  finally 
parallel  to  S4,  to  intersection  e  with  P5.  Through  this  last  point 
draw  a  line  parallel  to  the  last  ray  S5.  Now  S0  and  S5  are  com- 
ponents of  the  resultant  0  5  [Fig.  12  (a)~\  and  are  found  in 
proper  relative  position.  Produce  them,  therefore,  to  intersec- 
tion o '.  Through  this  point  the  resultant  must  pass.  Drawing 
then  through  o',  a  line  parallel  to  0  5,  we  have  the  resultant  in 
proper  position,  and  acting  in  the  direction  indicated  in  the  fig- 
ure, it  produces  equilibrium. 

Any  other  point  than  a,  upon  the  direction  of  P1}  assumed  as 
a  starting  point,  would  have  given  a  different  point  o' ;  so  also 
for  any  other  assumed  position  of  the  pole  C.  But  in  every 
case  we  shall  obtain  a  point  upon  the  line  of  direction  of  R^ 
already  found.  The  reader  may  easily  convince  himself  of  this 
by  making  the  construction  for  different  poles,  and  points  of 
beginning. 

ISTow  the  polygon  or  broken  line,  a  b  c  d  e,  we  call  the  equi- 
librium polygon — that  is,  it  is  the  position  which  a  system  of 
strings  or  struts,  S0  Si  S2,  etc.,  would  assume  under  the  action 
of  the  given  forces  at  the  assumed  points  of  application. 

Thus  Pt  acting  at  a,  is  held  in  equilibrium  by  the  forces  along 
S0  and  S1?  P2  acting  at  £,  by  Si  and  S2  and  so  on.  If  we  join 
any  two  points  in  the  line  of  direction  of  S0,  and  S5,  as  m  n  by 
a  line,  we  have  then  a  jointed  frame,  which  acted  upon  at  the 
apices  a.  .  .e  by  the  forces  P!.  .  .P5,  and  at  m  and  n  by  S0  and 
S5  is  in  equilibrium. 

For  S0  acting  at  m,  we  see  from  the  force  polygon  may  be 
replaced  by  a  force  a  0  parallel  and  opposed  to  the  resultant  R 
and  a  force  C  a  acting  along  the  line  L.  In  like  manner  S5  may 
be  replaced  by  a  C  and  5  a  parallel  and  opposed  to  the  result- 
ant. The  two  forces  a  C  and  C  a  being  equal  and  opposed 
balance  each  other  through  m  n,  while  the  sum  of  0  a  and  5  a 
is  equal  and  opposed  to  the  resultant  0  5.  There  is,  therefore, 
equilibrium,  and  m  and  n  may  be  considered  as  the  points  of 
support  of  the  frame  acted  upon  by  the  forces  Pt.  .  .P5  at  the 
apices  a.  .  .e,  a  Q  and  5  a  being  the  upward  reactions  at  the 
points  of  support. 

As  to  the  quality  of  the  strains  in  the  different  pieces ;  as 
before  the  reaction  at  m,  viz.,  a  0,  is  in  equilibrium  with  the 
strain  in  m  n  and  m  a.  Following  round,  then,  in  the  force 


24:  FOKCES    IN    THE    SAME    PLANE.  [CHAP.  II. 

polygon  from  a  to  0,  0  to  C  and  C  to  a,  and  referring  back  to 
the  frame,  we  find  strain  in  in  n  acting  towards  apex  ra,  there- 
fore compressive  ;  strain  in  m  a  acting  away  from  m,  theref ore 
tensile.  In  like  manner  Sx  S2  S3  are  in  tension,  while  S4  or  d  e 
and  S5  or  e  n  are  compressed. 

Hence  we  m&jfix  any  two  points  of  the  equilibrium  polygon 
by  joining  them  by  a  line.  The  forces  acting  at  these  points 
are  at  once  found  by  drawing  from  C  in  the  force  polygon  a 
parallel  to  this  line  to  intersection  with  resultant.  Thus  a  C 
(since  we  have  taken  m  n  parallel  to  Sx)  is  the  force  in  m  n  and 
a  0,  5  a,  are  the  forces  opposed  to  the  resultant  at  m  and  n. 

23.  Influence  of  a  Couple. — Among  the  forces  in  Fig.  12 
there  are  two,  P2  and  P3  which  are  equal,  parallel  and  opposite, 
the  direction  of  rotation  being  as  indicated  by  the  arrow.  Ex- 
amining the  equilibrium  polygon,  we  see  that  the  influence  of 
the  couple  is  to  shift  St  through  a  certain  distance  parallel  to 
itself,  to  S3.  Now  suppose  the  forces  composing  the  couple 
were  not  given,  but  the  value  of  the  couple  known,  from  the 
direction  of  rotation  and  the  area  of  the  triangle  A2  P2  P3, 
which  has  its  base  equal  to  one  of  the  forces  and  a  height  equal 
to  their  perpendicular  distance.  In  this  case  the  lines  1  2,  and 
S2  in  the  force  polygon,  would  disappear,  but  we  can  none  the 
less  find  the  point  d,  and  from  this  point  continue  the  polygon 
by  drawing  S4  and  S5,  and  thus  find  the  same  points  e  and  o'  as 
before.  To  do  this  we  have  simply  to  apply  the  principle 
deduced  in  Art.  21,  that  one  couple  can  be  replaced  by  another 
provided  the  area  of  the  triangle  is  constant. 

In  the  present  case  we  must  replace  the  given  couple  by 
another  whose  forces  are  Si  and  S3,  having  the  same  direction 
of  rotation. 

Lay  off  then  from  a,  a  i  equal  by  scale  to  S1  as  given  in  the 
force  polygon.  Describe  upon  St  the  triangle  a  g  h  equal  to 
the  given  area  A2  P2  P3.  Draw  g  i,  and  then  through  h,  h  7c 
parallel  to  g  i.  The  point  Jc  is  upon  the  line  of  direction  of 
S8,  or  in  other  words  the  area  of  the  triangle  i  J&  a  is  equal  to 
a  g  h.  The  proof  is  easy.  The  two  triangles  ig  hvxAigk 
are  equal,  since  they  have  the  same  base  i  g,  and  height.  But 
if  from  the  triangle  a  i  g  we  subtract  i  g  h,  we  obtain  a  g  h. 
If  from  the  same  triangle  a  i  g  we  subtract  i  g  k,  which  is  equal 
to  i  g  A,  we  obtain  i  k  a.  Equals  subtracted  from  equals  leave 
equals.  Hence  ik  a  is  equal  to  a  g  h. 


CHAP.  II.]  DIFFERENT   POINTS    OF   APPLICATION.  25 

If  then  through  7c  we  draw  a  line  parallel  to  S3  and  produce 
it  to  d.  we  have  the  same  point  as  before,  and  thus  from  d,  can 
continue  the  polygon. 

\_Note  that  the  direction  of  rotation  shows  the  side  of  Sj  upon 
which  the  point  7c  must  fall.  St  acts  away  from  a  [from  1  to 
C  in  (a)]  hence  for  rotation  as  shown  by  the  arrow,  g  must  fall 
above  S1?  and  $!  is  shifted  upwards. 

24.  Order  of  Forcc§  Immaterial. — As  in  the  case  of  a  com- 
mon point  of  application,  so  also  here,  the  order  in  which  the 
forces  are  laid  off  is  immaterial.     To  prove  this  for  two  forces 
is  sufficient,  as  by  continued  interchange  of  two  and  two,  we 
can  obtain  any  desired  order. 

Let  the  two  forces  be  P4  and  P5  (Fig.  13,  PL  3)  existing  either 
alone,  or  in  combination  with  others  preceding  and  following. 

Taking  the  forces  first  in  the  order  P4  P5,  we  have  the  equi- 
librium polygon  S3  S4  S5,  (b)  giving  the  point  a  in  the  result- 
ant. Taking  them  now  in  reverse  order,  P5  P4,  we  have  the 
polygon  S3  S'5  S'4  giving  the  same  point  a  in  the  resultant.  The 
resultant  in  the  force  polygon  (a),  viz.,  0  5,  is  of  course  un- 
changed in  intensity  and  direction  in  either  case.  It  is  required 
to  prove  that  in  the  second  case  the  last  string  S'4  is  not  only 
parallel  to  S5  in  the  first,  but  coincides  with  it. 

This  is  easy.  The  resultant  of  P4  P5  goes  through  a,  the  in- 
tersection of  S3  and  S5.  The  same  resultant  in  the  second  case 
must  also  pass  through  the  intersection  of  S3  and  S'4.  But  S3  is 
the  same  in  position  and  direction  in  both  cases.  If  the  second 
point  of  intersection  does  not  coincide  with  a,  still  it  must  lie 
somewhere  upon  S3.  Hence  as  the  resultant  must  pass  through 
both  points,  it  must  coincide  with  this  last  line ;  viz.,  S3.  But 
this  is  not  possible,  as  the  resultant  must  also  pass  through  d, 
the  point  of  intersection  of  the  forces,  or  when  these  do  not 
intersect  must  be  parallel  to  them.  As  therefore  S'4  must  be 
parallel  to  S5  (shown  by  the  force  polygon),  the  intersections  in 
each  case  must  coincide,  as  also  the  lines  S'4,  S5  themselves,  and 
the  polygon  from  e  on  has  the  same  course  in  either  case. 

25.  Pole  taken  upon  closing  line. — We  have  seen  (Art. 
20)  that  when  any  number  of  forces  are  in  equilibrium  both 
the  force  and  equilibrium  polygon  must  close.     There  is  one 
exception  to  this  statement.     Since  the  pole  may  be  taken  any- 
where, suppose  it  taken  somewhere  upon  the  line  closing  the 
force  polygon.     This  line,  as  we  know,  is  the  resultant,  and 


26  FORGES    IN    THE   SAME   PLANE.  [CHAP.  II. 

holds  the  other  forces  in  equilibrium.  But  now  the  equilibrium 
polygon  evidently  will  not  dose.  On  the  contrary  the  first  and 
last  strings  will  be  parallel.  This  position  of  the  pole  should 
then  be  avoided.  For  any  other  position  of  the  pole  our  rule 
holds  good ;  viz., 

If  the  force  polygon  closes  as  also  the  equilibrium  polygon, 
the  forces  are  in  equilibrium.  If  the  equilibrium  polygon 
however  does  not  close,  the  forces  cannot  be  replaced  by  a  single 
force  but  only  by  a  couple.  The  forces  of  this  couple  act  in 
the  parallel  end  lines  of  the  equilibrium  polygon,  and  are  given 
in  intensity  and  direction  of  action  by  the  line  from  the  pole 
to  the  beginning  of  the  force  polygon  [beginning  and  end  coin- 
ciding] . 

26.  Relation  between  two  equilibrium  polygons  with 
different  poles. — We  may  deduce  an  interesting  relation  be- 
tween the  two  equilibrium  polygons  formed  by  choosing  differ- 
ent poles,  with  the  same  forces  and  force  polygon. 

Thus  with  the  forces  P!  P2  P3  P4,  we  construct  the  force 
polygon  Fig.  14  (a),  PL  4.  Then  choose  a  pole  C  and  draw  SM, 
and  thus  obtain  the  corresponding  equilibrium  polygon  S0  a  b  c  d 
S4  Fig.  14  (b).  Choose  now  a  second  pole  C'.  Draw  S^  and 
construct  the  corresponding  polygon  S'0  a!  b'  c'  d'  S'4.  [In  our 
figure  c  and  c'  fall  accidentally  nearly  together.] 

Join  the  two  poles  by  a  line  CO'.  Then — any  two  corre- 
sponding strings  of  these  two  polygons  intersect  upon  the  same 
straight  line  M  N  parallel  to  C  C'.  Thus  S0  and  S'0  intersect 
at  g,  S'i  and  Sj  at  k,  S'2  and  S2  at  I,  S'8  and  S3  at  n,  S'4  and  S4  at 
m — and  all  these  points  g,  k,  I,  n  and  in,  lie  in  the  same 
straight  line  M  N  parallel  to  the  line  C  C'  connecting  the 
poles. 

The  proof  is  as  follows.*  If  we  decompose  Pt  into  the  com- 
ponents S0  Sj  and  S'0  S'^  these  components  are  given  in  inten- 
sity and  direction  by  the  corresponding  lines  in  the  force  poly- 
gon. If  we  take  the  two  first  as  acting  in  opposite  directions 
from  the  two  last,  they  hold  these  last  in  equilibrium.  The 
resultant  therefore  of  any  two  as  S0  and  S'0  must  be  equal  and 
opposed  to  that  of  the  remaining  two,  St  and  S'^  and  both  re- 
sultants must  lie  in  the  same  straight  line.  This  straight  line 
must  evidently  be  the  line  gk  joining  the  intersections  of  S0  S'0 

*  Ekmente  der  Graphischen  Statik.     Bauschinger.     Miinchen,  1871.    Pp. 
18-19. 


CHAP,  n.]  DIFFERENT   POINTS    OF    APPLICATION.  27 

and  Si  S\.  But  from*  the  force  polygon  we  see  at  once  that  the 
resultant  of  S0  and  S'0  is  given  in  direction  and  intensity  by 
C  C',  and  this  is  also  the  resultant  of  Si  and  S\.  The  line  join- 
ing g  and  k  must  therefore  be  parallel  to  C  C'.  For  the  second 
force  P2  we  can  show  similarly  that  the  line  joining  k  and  I  is 
parallel  to  C  C'.  But  k  is  a  common  point  of  both  lines — hence 
g  k  and  I  lie  in  the  same  straight  line  parallel  to  C  C'. 

[NOTE. — The  pure  geometric  proof  is  as  follows  :  The  two 
complete  quadrilaterals  0  1  C'  C  and  g  k  a'  a  have  five  pairs  of 
corresponding  sides  parallel,  viz.,  0  1  and  a  a',  a  a'  1  C'  and 
a!  k,  C  0  and  a  g,  o  C'  and  a'  g,  1  C'  and  ak'  hence  the  sixth 
pair  are  also  parallel,  viz.,  C  C'  and  g  k.  In  like  manner  for 
1  2  C  C'  and  lkbr  b  and  so  on.~\ 

We  can  make  use  of  this  principle  in  order  from  one  given 
equilibrium  polygon  S0  a  b  c  d  S4  and  pole,  to  construct  another, 
the  direction  of  C  C'  being  known.  For  this  purpose,  having 
assumed  the  position  of  the  first  string  S'0  we  draw  through  its 
intersection  g  with  S0  a  line  M  N  parallel  to  C  C'.  The  next 
string  must  therefore  pass  through  the  intersection  a'  of  S'0  and 
P!  and  through  the  point  k,  of  intersection  of  the  second  string 
of  the  first  polygon  and  the  line  M  N.  It  is  therefore  deter- 
mined. The  next  side  must  pass  through  V  and  Z,  and 
so  on. 

[Note.  Observe  that  the  intersections  r  and  r'  of  the  first  and 
last  lines  of  both  polygons  must  lie  in  a  straight  line  parallel  to 
0  4,  the  direction  of  the  resultant.] 

27.  Mean  polygon  of  equilibrium.— Since  the  pole  may 
have  any  position,  let  us  suppose  it  situated  in  one  of  the  angles 
of  the  force  polygon.  It  is  evident  that  the  first  line  of  the 
corresponding  equilibrium  polygon,  then  coincides  with  the  first 
force.  If  now  the  pole  be  taken  at  the  beginning  of  the  first 
force  in  the  force  polygon,  then  the  first  side  of  the  correspond- 
ing equilibrium  polygon  will  coincide  with  the  first  force,  and 
the  last  line  will  be  the  resultant  itself  in  proper  position. 

Take  for  instance,  the  pole  at  o  in  the  force  polygon,  Fig.  15 
(a\  PL  4.  The  first  side  S0  reduces  to  zero.  The  next  St  coin- 
cides with  0  1.  In  (b)  'therefore  Pt  is  the  first  side  of  the  equi- 
librium polygon.  The  next  side  S2  corresponds  with  S2  in  (a). 
Thus  we  obtain  the  polygon  a  b  c  d  e,  the  last  side  of  which  S7, 
is  the  resultant  itself.  That  is,  S2  is  the  resultant  of  Pt  and  P2, 
S:>,  of  P^,  S4  of  P14  and  so  on.  Every  line  in  the  polygon  then 


28  FORCES    IN    THE    SAME   PLANE.  [CHAP.  II. 

is  the  resultant  of  the  forces  preceding,  and  we  call  such  a 
polygon  the  mean  polygon  of  equilibrium. 

If  we  wish  to  find  the  mean  polygon  for  P^  we  have  only  to 
take  the  new  pole  C'  at  2  in  the  force  polygon  (a).  According 
to  the  preceding  Art.,  each  side  of  the  new  polygon  must  pass 
through  the  intersection  of  the  corresponding  side  of  the  first 
with  the  line  S2  which  passes  through  a  and  is  parallel  to  C  C'. 
Thus  S'4  must  pass  through  ~b'  and  o.  S'5  through  d  and  n,  and 
so  on.  S'7  is  the  resultant  of  PS7,  and  since  S2  is  the  resultant 
of  P^;  S7,  the  resultant  of  P1Jr,  must  pass  through  the  intersec- 
tion m  of  S'7  and  S2. 

We  observe  here  again  the  influence  of  the  couple  P5  and  P6. 
S4  and  S  4  are  simply  shifted  through  certain  distances,  without 
change  of  direction,  to  S6  and  S'6 ;  and  as  we  have  seen  above, 
knowing  the  direction  of  rotation,  and  the  moment  of  the  couple, 
we  might  have  omitted  it  in  the  force  polygon  and  still  obtained 
S7  and  S'7  as  before. 

2§.  Line  of  pressures  in  an  arch. — The  practical  applica- 
tion of  the  above  will  be  at  once  seen  in  the  consideration  of 
an  arch.  Thus  with  the  given  horizontal  thrust  applied  at  a 
given  point  of  the  arch,  and  the  forces  Pw,  we  construct  the 
force  polygon  C  o  5,  and  then  the  line  of  pressures  abed. 
[Fig.  16,  PI.  4.] 

Required  with  another  thrust  H'  =  o  C'  acting  at  another 
point,  and  the  same  forces  Pw,  to  construct  the  corresponding 
line  of  pressures.  To  do  this  we  have  only  to  lay  off  o  C'  equal 
to  the  new  horizontal  thrust,  then  choose  a  point  of  the  force 
line,  as  3,  as  a  pole  and  draw  the  corresponding  polygon, 
Jc  op  Jc  •  the  point  of  intersection,  &,  is  a  point  upon  the  line 
m  n  parallel  to  o  C,  and  upon  this  line  will  be  found  the  inter- 
section of  corresponding  sides  of  the  two  polygons.  Thus  from 
the  intersection  of  the  side  ap  of  the  first  polygon  with  m  n, 
draw  a  line  to  o  and  we  have  a'.  From  the  intersection  ~b  of 
the  second  line  of  the  first  polygon  draw  a  line  to  a',  and  we 
have  b'  #',  and  so  on. 

29. — The  preceding  articles  comprise  all  the  most  important 
principles  of  the  Graphical  Method  which  can  be  deduced  in- 
dependently of  its  practical  applications.  Future  principles 
will  be  best  demonstrated,  and  at  the  same  time  illustrated,  by 
considering  the  various  special  applications  of  the  method,  and 
to  these  applications  we  shall  therefore  now  proceed. 


CHAP.  III.]  CENTRE    OF    GRAVITY.  29 


CHAPTER   III. 


CENTRE   OF   GRAVITY. 

30.  General  Method. — One  of  the  most  obvious  applica- 
tions of  the  new  method  as  thus  far  developed,  is  to  the  deter- 
mination of  the  centre  of  gravity  of  areas  and  solids.  We 
shall  confine  ourselves  to  areas  only,  merely  observing  that  all 
the  principles  hitherto  developed  apply  equally  well  to  forces 
in  space.  The  forces  being  given  by  their  orthographic  pro- 
jections upon  two  planes  after  the  manner  of  descriptive  geo- 
metry, the  projections  upon  each  plane  may  be  dealt  with  as 
forces  lying  in  that  plane,  and  thus  the  projections  of  the  force 
and  equilibrium  polygons,  the  resultant,  etc.,  determined. 

A  body  under  the  action  of  gravity  may  be  considered  as  a 
body  acted  upon  by  parallel  forces.  The  resultant  of  these 
forces  being  found  for  one  position  of  the  body  [or  the  body 
being  considered  as  fixed,  for  one  common  direction  of  the 
forces]  may  have  its  point  of  application  anywhere  in  its  line 
of  direction. 

For  a  new  position  of  the  body  [or  another  direction  of  the 
forces]  there  is  another  position  for  the  resultant.  Among  all 
the  points  which  may  be  considered  as  points  of  application  of 
these  two  resultants  there  is  one  which  remains  unchanged  in 
position,  whatever  the  change  in  direction  of  the  parallel  forces. 
This  point  must  evidently  lie  upon  all  the  resultants,  and  is 
therefore  given  by  the  intersection  of  any  two. 

It  is  hardly  necessary  to  give  illustrations  of  the  method  of 
procedure. 

Generally,  we  divide  up  the  given  area  into  triangles,  trapez- 
oids,  rectangles,  etc.,  and  reduce  the  area  of  each  of  these  fig- 
ures to  a  rectangle  of  assumed  base.  The  heights  of  these 
reduced  rectangles  will  then  be  proportional  to  the  areas,  and 
hence  to  the  force  of  gravity  acting  upon  them ;  i.e.,  to  their 
weights.  Consider  then  these  heights  as  forces  acting  at  the 
centres  of  gravity  of  the  partial  areas.  Construct  the  force 


30  CENTRE   OF    GRAVITY.  [CHAP.  III. 

polygon  by  laying  them  off  one  after  the  other.  Choose  a  pole 
and  draw  lines  from  it  to  the  beginning  and  end  of  each  force. 
These  lines  will  give  the  sides  of  the  funicular  or  equilibrium 
polygon.  Anywhere  in  the  plane  of  the  figure,  draw  a  line 
parallel  to  the  first  of  these  pole  lines  (S0).  Produce  it  to  inter- 
section with  the  first  force  (P^,  prolonged  if  necessary.  From 
this  intersection  draw  a  parallel  to  the  second  pole  line  (S^,  and 
produce  to  intersection  with  second  force  (P2).  So  on  to  last 
pole  line,  which  produce  to  intersection  with  first  pole  line. 
Through  this  point  the  resultant  must  pass,  and  of  course  it 
must  be  parallel  to  the  forces. 

Now  suppose  the  parallel  forces  all  revolved  say  90°,  the 
points  of  application  remaining  the  same.  Evidently  the  new 
force  polygon  will  be  at  right  angles  to  the  first,  as  also  the 
new  pole  lines,  each  to  each.  It  is  unnecessary  then  to  form 
the  new  force  polygon.  The  directions  of  the  new  pole  lines 
are  given  by  the  old,  and  this  is  all  that  is  needed. 

Anywhere  then  in  the  plane  of  the  figure,  draw  a  line  (S'0) 
perpendicular  to  the  first  pole  line  (S0)  previously  drawn,  and 
prolong  to  intersection  with  new  direction  of  first  force  (P/). 
Through  this  point  draw  a  perpendicular  (S/)  to  second  pole 
line,  to  intersection  with  new  direction  of  second  force  (P2') 
and  so  on.  We  thus  find  a  point  for  new  resultant,  parallel  to 
new  force  direction.  Prolong  this  resultant  to  intersection 
with  first  and  the  centre  of  gravity  is  determined. 

[NOTE. — If  the  area  given  has  an  axis  of  symmetry,  that  can 
of  course  be  taken  as  one  resultant,  and  it  is  then  only  necessary 
to  make  one  construction  in  order  to  find  the  other.] 

The  given  area  of  irregular  outline  must,  as  remarked  above, 
be  divided  by  parallel  sections  into  areas  so  small  that  the  out- 
lines of  these  areas  may  be  considered  as  practically  straight 
lines.  The  forces  are  then  taken  as  acting  at  the  centres  of: 
gravity  of  these  areas.  This  division  will  give  us  generally  a 
number  of  triangles  and  trapezoids. 

It  is  therefore  desirable  to  reduce  graphically  to  a  common 
base  the  area  of  these  triangles  and  trapezoids,  and  for  this  pur- 
pose the  following  principles  will  prove  of  service : 

32.  Reduction  of  Triangle  to  equivalent  Rectangle  of 
given  Base. — Let  b  be  the  base  and  h  the  height.  Then  area 

=  — .     Take  a  as  the  given  reduction  base,  and  let  x  represent 


CHAP.  III.]  CENTRE   OF    GRAVITY.  31 

the  height  of  the  equivalent  rectangle.     Then 

bfi      k        x 

ax  =  —  or  —  =  — -. 
2,       a       %b 

Now  a,  £,  and  h  being  given,  it  is  required  to  find  x  graphi- 
cally. 

Let  A  B  C  be  the  triangle,  and  D  the  middle  of  the  base. 
[Fig.  17,  PL  5.]  Lay  off  A  E  =  h  and  A  F  =  a.  Draw  F  D, 
and  parallel  to  F  D  draw  E  x.  Then  A  a?  is  the  required 
height. 

-r,        A  a?      A  E        x        h 
For:AD  =  AF°rp  =  a 

As  to  the  centre  of  gravity  of  the  triangle,  it  is  evidently  at 
the  intersection  of  the  lines  from  each  apex  to  the  centre  of 
the  opposite  side ;  since  these  lines  are  axes  of  symmetry. 

33.  Reduction  of  Trapezoid  to  equivalent  Rectangle. 

— In  the  trapezoid  A  B  C  D,  Fig.  18,  PL  5,  draw  through  the 
middle  points  of  A  D  and  B  C  perpendiculars  to  D  C,  and  pro- 
duce to  intersections  E  and  F  with  A  B  produced. 

Then  lay  off  F  g  —  a  —  the  given  reduction  base,  and  draw 
g  E  intersecting  D  C  in  a?.  Then  H  x  is  the  required  height. 

T,     EF       H  x      EF         x 

H  Yyp (\Y  • 

F^~HE         a     ~HE' 

hence  ##=:EFxHEi=  area. 

To  find  the  centre  of  gravity ,  draw  a  line  through  the  mid- 
dle points  of  the  parallel  sides  A  B  and  D  C.  This  line  is  an 
axis  of  symmetry.  Prolong  A  B  and  C  D  and  make  C  a  — 
A  B  and  A  b  =  C  "D  and  join  a  and  ~b.  Then  the  intersection 
of  a  ~b  with  the  axis  of  symmetry  gives  the  centre  of  gravity. 

The  construction  for  the  reduction  of  a  parallelogram  is  pre- 
cisely similar.  [Fig.  18  (5).] 

The  points  F  and  E  here  coincide  with  A  and  B,  and  we 
have 

A  x      AB 

T-  =  = — ,  or  #  #  —  h  x  A  B  =:  area. 

h,  B    g 

The  same  construction  also  holds  good,  of  course,  for  a  rect- 
angle or  square.  The  centre  of  gravity  in  each  case  is  at  the 
intersection  of  two  diameters,  since  these  are  axes  of  symmetry. 

31.  Reduction  off  Quadrilaterals  Generally. — In  general 


32  CENTRE  OF  GRAVITY.  [CHAP.  III. 

any  quadrilateral  may  be  divided  into  two  triangles  which  may 
be  reduced  separately,  or  into  a  triangle  and  trapezoid. 

It  is  also  easy  to  reduce  any  quadrilateral  to  an  equivalent 
triangle,  which  may  then  be  reduced  by  Art.  32  to  an  equiva- 
lent rectangle  of  given  base. 

Thus  we  reduce  the  quadrilateral  A  B  C  D  [Fig.  18  (c)] 
to  an  equivalent  triangle  by  drawing  C  Cx  parallel  to  D  B  to 
intersection  Ct  with  A  B,  and  joining  Ct  and  D.  The  triangle 
D  B  Ci  is  then  equal  to  D  B  C,  and  hence  the  area  A  D  Cx  is 
equal  to  A  B  C  D.  The  triangle  A  D  C^  can  now  be  reduced 
to  an  equivalent  rectangle  of  given  base  by  Art.  32. 

The  centre  of  gravity  of  the  quadrilateral  may  be  found  as 
follows : 

Draw  the  diagonals  A  C  and  B  D  and  mark  the  intersection 
E.  Make  A  Ex  =  C  E  and  B  E2  =  D  E,  also  find  the  centres 
Q!  and  O2  of  the  diagonals  A  C  and  B  D.  Join  O2  Et  and  Ol 
E2 ;  the  intersection  S  of  these  two  lines  is  the  centre  of  gravi- 
ty required. 

The  above  is  sufficient  to  enable  us  to  find  the  centre  of  gravity 
of  any  given  area  of  regular  or  irregular  outline.  The  method 
may  be  applied  to  finding  the  centre  of  gravity  of  a  loaded 
water-wheel  (as  given  in  Der  Constructeur,  Reuleaux,  Art.  47), 
and  many  similar  problems.  The  reader  will  have  no  difficul- 
ty, following  the  general  method  indicated  in  Art.  30,  in  mak- 
ing such  applications  for  himself.  The  method  itself  is  so  sim- 
ple that  it  is  unnecessary  to  give  here  any  practical  examples 
in  illustration.  We  shall,  moreover,  have  occasion  to  return  to 
the  subject  in  the  consideration  of  moment  of  inertia  of  areas. 

"We  pass  on  therefore  to  the  moment  of  "rotation  of  forces  in 
a  plane. 


CHAP.  IV.]       MOMENT  OF  ROTATION  OF  FORCES. 


CHAPTEE    IY. 


MOMENT    OF    ROTATION    OF    FORCES    IN    THE    SAME    PLANE. 

35.  The  "Moment"  of  a  Force  about  any  Point  is  the 

product  of  the  force  into  the  perpendicular  distance  from  that 
point  to  the  line  of  direction  of  the  force.  The  importance 
and  application  of  the  "  moment "  in  the  determination  of  the 
strains  in  the  various  pieces  of  any  structure  will  be  evident  by 
referring  to  Art.  14,  where  Ritter's  "  method  of  sections  "  is 
alluded  to.  In  general,  when  the  moments  of  all  the  exterior 
forces  acting  upon  a  framed  structure  are  known,  the  interior 
forces,  or  the  strains  in  the  various  pieces,  can  be  easily  ascer- 
tained. 

As  we  shall  immediately  see,  these  moments  are  given 
directly  in  any  case  by  the  "  equilibrium  polygon" 

36.  Ciilmann's  Principle. — If  a  force  P  be  resolved  into 
two  components  in  any  directions  as  b  C,  b  C^  (Fig.  19,  PL  5), 
and   these  components   be  prolonged,  it  is   evident  that   the 
moment  of  P  with  reference  to  any  point  as  a  situated  any- 
where in  the  line  c  d  parallel  to  P,  is  P  x  b  a.     But  if  from  C 
we  draw  the  perpendicular  H  to  P,  then  by  similar  triangles, 

P  :  H  ; :  c  d  :  I  a  ; 

Pxb  a  =  Hxc  d. 

That  is,  the  moment  ofP  with  respect  to  any  point  a  is  equal 
to  a  certain  constant  H  multiplied  by  the  ordinate  c  d,  paral- 
lel to  P  and  limited  by  the  components  prolonged.  The  con- 
stant H  we  call  the  "pole  distanced 

This  holds  good  for  any  point  whatever,  and  we  have  only  to 
remember  that  if  we  assume  the  ordinates  to  the  right  of  P  as 
positive,  those  to  the  left  are  negative. 

"We  can  choose  the  pole  C  where  we  please,  and  thus  obtain 
various  values  for  H,  but  for  any  one  value  the  corresponding 
ordinates  are  proportional  to  tlie  moments. 


34  MOMENT   OF   KOTATION    OF   FOKOES.  [CHAP.  IV. 

The  above  principle  is  due  to  Culmann,  and  will  be  referred 
to  hereafter  as  Culmann' s  principle. 

37.  Application  of  the  above  to  Equilibrium  Polygon. 

—Let  P14  be  a  number  of  forces  given  in  position  as  repre- 
sented in  Fig.  19  (a)  PL  5.  By  forming  foa  force  polygon  Fig. 
19  (b),  choosing  a  pole  C,  and  drawing  S0  S1?  S2,  etc.,  we  form 
the  equilibrium  polygon  abed  ef,  Fig.  19  (a). 

The  resultant  of  the  forces  P14  acts  in  the  position  and  direc- 
tion given  in  the  Fig.  How,  as  we  have  seen  in  Art.  22, 
regarding  the  broken  line  a  b  c  d  e  as  a  system  of  strings,  we 
may  produce  equilibrium  by  joining  any  two  points  as  a  and/" 
by  a  line,  and  applying  at  a  and  f  the  forces  S0  and  S4.  Let  us 
suppose  this  line  a  f  perpendicular  to  the  direction  of  the 
resultant.-  Since  we  can  suppose  the  broken  line  or  polygon 
fastened  at  any  two  points  we  please,  this  is  allowable,  and 
does  not  affect  the  generality  of  our  conclusion. 

Then  the  compression  in  the  line  a  f  is  given  by  H,  the 
"pole  distance"  or  the  distance  of  the  pole  C  from  the  result- 
ant in  the  force  polygon.  We  have  therefore  at  a  the  force 
H  and  Vi  =  H  0  acting  as  indicated  by  the  arrows.  At  a  then 
Y!  acting  up,  H  and  S0  acting  away  from  a,  are  in  equilibrium, 
or  Vx  is  decomposed  into  H  and  S0,  as  shown  by  the  force 
polygon. 

According  to  Culmanrts  principle  then,  the  moment  of  Vt 
with  reference  to  any  point,  as  m  or  o,  is  equal  to  H  x  o  m. 
Therefore  H  being  known,  the  ordinates  between  a  f  and  S0 
are  proportional  to  the  moment  of  Vt  at  any  point.  Vx  acting 
upwards  gives  positive  rotation  (left  to  right)  with  respect 
to  m. 

At  the  point  b,  Pt  may  be  replaced  by  a  force  0  K  parallel  to 
R  and  a  force  K  1  along  St  [see  force  polygon].  This  we  see 
at  once  from  the  force  polygon  where  0  K  and  K  1  make  a 
closed  polygon  with  P1?  and  taken  as  acting  from  0  to  K  and 
K  to  1,  replace  Plt  But  these  two  forces  are  in  equilibrium 
with  Si  and  S0,  or  1  C  and  C  0  [see  force  polygon],  and  since 
K  1  and  1  K  balance  each  other,  all  the  forces  acting  at  b  may 
be  replaced  by  S0,  0  K  and  K  C.  We  have  then  at  b  the  force 
0  K  resolved  into  components  in  the  directions  S0  and  S^ 

By  Culmanrfs  principle,  therefore,  the  moment  of  O  K 
about  any  point  as  m,  is  proportional  to  the  ordinate  n  m,  and 
since  0  K  acts  downward  this  moment  is  negative.  Hence  the 


CHAP.  IV.]       MOMENT  OF  ROTATION  OF  FOKCES.  35 

resultant  moment  at  m  or  o  of  the  components  at  a  and  b  par- 
allel to  R,  is  proportional  to  the  ordinate  o  n. 

So  for  any  point,  the  ordinate  included  by  the  polygon  ab  c 
d  ef,  and  the  dosing  line  af,  to  the  scale  of  length  multiplied 
by  the  "pole  distance  "  H  to  the  scale  of  'force ;,  gives  the  mo- 
ment at  that  point  of  the  components  parallel  to  the  resultant. 

The  practical  importance  and  application  of  this  principle 
will  appear  more  clearly  in  the  consideration  of  parallel  forces 
in  the  next  Chapter. 


36  MOMENT   OF   RUPTURE   OF   PARALLEL   FORCES.    [CHAP.    V. 


CHAPTEE   Y. 


MOMENT   OF   RUPTURE   OF   PARALLEL   FORCES, 

3§.  Equilibrium  Polygon. — Since  the  forces  acting  upon 
structures  are  generally  due  to  the  action  of  gravity,  these 
forces  may  be  considered  as  parallel  and  vertical,  and  in  all 
practical  cases  therefore,  we  have  to  do  with  a  system  of  paral- 
lel forces. 

Given  any  number  of  parallel  forces  P^,  PL  6,  Fig.  20 ; 
required  to  find  the  direction,  intensity  and  position  of  the 
resultant,  and  the  moment  of  rotation  at  any  point. 

1st.  Draw  the  force  polygon,  in  this  case  it  is,  of  course,  a 
straight  lino. 

2d.  Choose  a  pole  C,  and  draw  the  lines  S0.  S1?  S2,  etc. 

3d.  Draw  the  string  or  equilibrium  polygon  a  b  c  d  e  f. 
Considering  this  polygon  as  a  system  of  strings,  the  forces  will 
be  held  in  equilibrium  if  wre  join  any  two  points,  as  a  and  g^ 
by  a  strut  or  compression  piece,  and  apply  at  a  and  g  the  up- 
ward forces  Vj  and  V2. 

4th.  Prolong  a  1}  and  f  g  to  their  intersection  o.  Through 
this  point  the  resultant  must  pass.  It  is  of  course  parallel  and 
equal  to  the  sum  of  the  forces. 

Now,  if  a  g  is  assumed  horizontal,  the  perpendicular  H  to 
the  force  line,  or  the  "pole  distance"  divides  the  resultant  0  5 
into  the  two  reactions  Y!  and  V2  (Art.  22). 

All  the  forces  in  the  equilibrium  polygon  have  the  same 
horizontal  projection  H,  in  the  force  polygon. 

Let  a  g  represent  a  beam  resting  upon  supports  at  a  and  g. 
We  have  then  at  once  the  vertical  reactions  V1  and  V2  or  ~k  0 
and  5  &,  which,  in  order  to  cause  equilibrium,  must  act  up- 


For  the  moment  at  any  point,  as  0,  due  to  V1?  we  have,  by 
Culmann's  principle,  m  o  multiplied  by  H.  The  triangle  formed 
by  a  5,  a  g,  and  Pl5  gives  then  the  moment  of  rupture  at  any 
point  of  the  beam  as  far  as  Pj.  For  a  point  o,  beyond  Pj,  the 


CHAP.  V.]     MOMENT   OF   RUPTURE    OF   PARALLEL   FORCES.  37 

moment  due  to  V1?  must  be  diminished  by  that  due  to  P1?  since 
these  forces  act  in  opposite  directions,  and  rotation  from  left 
to  right  upon  the  left  of  any  point  is  considered  positive.  We 
see  at  once  from  the  force  polygon  that  Pt  is  resolved  into  S0 
and  Sj  or  into  a  b  and  b  c.  Hence  the  moment  at  o  due  to  P! 
is  m n  multiplied  by  H.  The  total  moment  at  o  is  then  mo  — 
m  n  =  n  o,  multiplied  by  H. 

Hence  we  see  that  the  ordinates  to  the  equilibrium  polygon 
from  the  closing  line  a  g,  are  proportional  to  the  total  mo- 
ments ;  while  the  ordinate  at  any  point  between  any  two  adja- 
cent sides  of  this  polygon,  prolonged,  represents  the  moment  at 
that  point  of  a  force  acting  in  the  vertical  through  the  inter- 
section of  these  two  sides. 

[The  reader  should  make  the  construction,  changing  the  order  in  which 
the  weights  are  taken,  and  thus  satisfy  himself  that  the  order  is  a  matter ' 
of  indifference.  As  to  the  direction  of  the  reactions  Vi,  Va,  it  must  be 
remembered  that  a  &  is  to  be  replaced  by  Vi  and  H,  hence  Vi  must  be  op- 
posed to  C  0,  the  direction  obtained  by  following  round  in  the  force  poly- 
gon the  triangle  0  1  O.  Force  and  distance  scales  should  also  be  assumed. 
Thus  the  ordiuates  to  the  equilibrium  polygon  scaled  off  say  in  inches,  and 
multiplied  by  the  number  of  tons  to  one  inch,  and  then  by  the  "pole  dis- 
tance "  taken  to  the  assumed  scale  of  distance,  will  give  the  moments  of 
any  point.] 

The  resultant  of  any  two  or  more  forces  must  pass  through 
the  intersection  of  the  outer  sides  of  the  equilibrium  polygon 
for  those  forces  (Art.  16).  Thus,  the  resultant  of  Pt  arid  P3 
must  pass  through  the  intersection  of  a  b  and  c  d.  Of  Vt  and 
Pb  through  the  intersection  of  a  g  and  b  c ;  of  Pt  P2  and  P3, 
through  intersection  of  a  b  and  d  e,  and  so  on.  In  every  case 
the  intensity  and  direction  of  action  of  the  resultant  is  given 
directly  by  simple  inspection  of  the  force  polygon. 

Thus  from  the  force  polygon  we  see  that  the  resultants  Tc  2 
and  Jc  3  of  Vj.  P!  P2  and  Vx  P!  P2P3,  act  in  different  directions. 
Their  points  of  application  are  at  the  intersection  of  c  d  and  d  e 
respectively  with  a  g,  or  upon  either  side  of  d  in  the  equilibrium 
polygon.  At  d  the  ordinate  and  hence  the  moment  is  greatest, 
and  at  this  point  the  tangent  to  the  polygon  is  parallel  to  a  g. 
If  we  had  a  continuous  succession  of  forces ;  if  a  g,  for  in- 
stance, were  continuously  or  uniformly  loaded ;  the  equilibri- 
um polygon  would  become  a  curve,  and  the  tangent  at  d  would 
then  coincide  with  the  very  short  polygon  side  at  that  point. 


38  MOMENT   OF   RUPTURE    OF   PARALLEL   FORCES.     [CHAP.  V. 

The  points  of  application  of  the  resultants  of  all  the  forces 
right  and  left  of  d  are  then  at  the  intersection  of  this  tangent 
with  a  g,  or  at  an  infinite  distance. 

At  d  then  we  have  a  couple,  the  resultant  of  which  is  as  we 
have  seen  (Art.  20),  an  indefinitely  small  force  acting  at  an 
indefinitely  great  distance.  That  is,  with  reference  to  d,  the 
forces  acting  right  and  left  cannot  be  replaced  by  a  single 
force. 

Hence  generally  :  at  the  point  of  maximum  moment  ("  cross 
section  of  rupture"),  the  resultant  of  the  outer  forces  on  either 
side  reduces  to  an  indefinitely  small  and  distant  force,  the 
direction  of  which  is  reversed  at  this  point,  and  the  point  of 
application  of  which  changes  from  one  side  to  the  other  of  the 
equilibrium  polygon.* 

The  "  cross  section  of  rupture  "  then,  is  that  point  where  the 
weight  of  that  portion  of  the  girder  between  it  and  the  end  is 
equal  to  the  reaction  at  that  end,  or  where  the  resultant  changes 
sign. 

The  value  of  the  moment  at  this  point,  is  therefore  equal  to 
the  product  of  the  reaction  at  one  end  into  its  distance  from 
the  point  of  application  of  the  equal  resultant  of  all  the  loads 
between  that  end  and  the  point. 

Thus  for  a  beam  uniformly  loaded  with  w  per  unit  of  length, 

the  reaction  at  each  end  is  -^-*  From  the  above,  the  cross  sec- 
tion of  rupture  is  then  at  the  middle.  The  point  of  application 
of  the  resultant  of  the  forces  acting  between  one  end  and  the 

A  I    ,  .  w  I     I      w  Z2 

middle  is  at  -r,  hence  the  maximum  moment  is-^-  x~:  =  -5-. 

4:  A  4:  O 

39.  Beam  with  Two  Equal  and  Opposite  Forces  beyond 
the  Supports. — The  ordinates  to  the  equilibrium  polygon  thus 
give,  as  it  were,  a  picture  or  simultaneous  view  of  the  change 
and  relative  amount  of  the  moments  at  any  point.  The  point 
where  the  moment  is  greatest,  i.e.,  where  the  beam  is  most 
strained,  is  at  once  determined  by  simple  inspection. 

Let  us  take  as  an  example  a  beam  with  two  equal  and  oppo- 
site forces  leyond  the  supports.  Thus,  Fig.  21,  PL  6,  suppose 
the  beam  has  supports  at  A  and  B,  the  forces  being  taken  in 
the  order  as  represented  by  P!  P2.  We  first  construct  the  force 

*  Die  GrapMscJie  Statik.—  Culmann,  p.  127. 


CHAP.  V.]  MOMENT  OF  RUPTURE  OF  PARALLEL  FORCES.         39 

polygon  from  0  to  1,  and  1  to  2  or  0.  Next  choose  a  pole  C, 
and  draw  S0  Si  and  S2.  Draw  then  a  parallel  to  S0  till  intersec- 
tion with  first  force,  Pt,  then  parallel  with  Sj  to  second  force, 
P2,  then  parallel  to  S2  or  S0  to  intersection  with  vertical  through 
support  B,  and  finally  draw  the  closing  line  L.  A  line  through 
C,  parallel  to  L,  gives  as  before  the  vertical  reactions.  Follow- 
ing round  the  force  polygon,  we  find  at  A  the  reaction  down- 
wards, since  S0  acts  from  C  to  0  and  is  to  be  replaced  (Art.  4) 
by  L  and  Vx ;  at  B  reaction  upwards,  since  P2  acts  up,  and  fol- 
lowing round,  S2  acts  from  0  to  C.  .Both  reactions  are  equal  to 
a  0.  At  A  then  the  support  must  be  above,  and  at  B  below  the 
beam.  The  shaded  area  gives  the  moments  to  pole  distance  H. 
Had  we  taken  the  pole  in  the  perpendicular  through  o,  S0  would 
have  been  parallel  with  the  beam  itself.  This  is,  however,  a 
matter  of  indifference.  The  moment  area  may  lie  at  any  in- 
clination to  the  beam.  We  also  see  here  again  the  effect  of  a 
couple  (Art.  23).  S0  is  simply  shifted  through  a  certain  distance 
to  S2,  parallel  to  S0,  and  therefore  the  moment  at  any  point  be- 
tween P2  and  B  is  constant.  This  is  generally  true  of  any 
couple,  as  we  have  already  seen,  Article  21,  and  may  be  proved 
analytically  as  follows : 

Let  the  distance  between  the  forces  be  a  =  A  B,  Fig.  22. 
Then  for  any  point  o,  we  have  P  X  (&  +  B  6>)— PxBo  —  P  [a  +  ^B  o 
— B  o]  =  P  a.  For  o'  between  A  and  B,  P  x  A  o'+P  x  o'  B= 
P  [A  </  +  <?' B]  =Pa. 

So  also  for  any  point  to  the  left,  the  same  holds  true. 

Graphically  the  proof  is  as  follows : 

Decompose  both  forces  into  parallel  components,  Fig.  23. 
Then  for  any  point,  as  o,  we  have  the  moment  M  =  H  x  m  n— 
Hxmp  or  M  =  —  H  x  n  p.  But  n  p  is  the  constant  ordinate 
between  the  parallel  components  A  n  and  A  p. 

We  see,  therefore,  by  simple  inspection,  that  the  distance  of 
P!  and  P2  from  the  support  B,  Fig.  21,  has  no  influence  what- 
ever upon  the  moment  or  strain  in  A  B,  provided  the  distance 
between  the  points  of  application  remains  the  same,  and  that 
the  moment  at  all  points  between  P2  and  the  support  B  is  con- 
stant and  a  maximum.  From  B  and  P2  the  moments  decrease 
left  and  right,  and  become  zero  at  A  and  Pj. 

1O.  Beam  with  Two  Equal  and  Opposite  Forces  be- 
tween the  Two  Supports. — Let  the  beam  A  B,  Fig.  24,  PL  6, 
be  acted  upon  by  the  two  equal  and  opposite  forces  P1  P2. 


4:0  MOMENT   OF  KUPTURE   OF   PARALLEL   FORCES.      [CHAP.  V. 

Construct  the  force  polygon  012.  Choose  a  pole  C  and  draw 
C  0,  C  1,  C  2.  Parallel  to  C  0,  draw  the  first  side  of  the  equi- 
librium polygon  to  intersection  with  first  force  Pt;  then  paral- 
lel to  C  1  to  second  force  P2 ;  then  parallel  to  C  2  to  d.  Join  d 
and  0.  Parallel  to  this  draw  C  a  in  force  polygon.  Then  0  a 
is  the  vertical  reaction  at  A,  which  acts  upwards,  since  it  niust 
with  C  a  replace  C  0 ;  and  C  0,  when  we  follow  round  from  o  to 
1  and  1  to  C,  acts  from  C  to  0. 

We  have  the  same  vertical  reaction  at  B,  but  here,  since  we 
must  follow  from  1  to  2  and  2  to  C,  C  2  acts  from  2  to  C,  hence 
following  round,  the  reaction  at  B  is  downward.  The  shaded 
area  gives  the  moments  to  pole  distance  H,  as  before. 

We  see  at  once  that  at  a  certain  point  e  the  moment  is  zero. 
Left  and  right  of  this  point  the  moment  is  positive  and  nega- 
tive. At  the  point  itself  we  have  a  point  of  inflection,  and 
here,  since  the  moment  is  zero,  there  is  no  longitudinal  strain. 
At  jb  and  c  the  moments  are  greatest ;  here  the  beam  is  most 
strained,  and  at  these  points,  therefore,  are  the  "  cross  sections 
of  rupture."  Here  again,  if  we  had  taken  the  pole  C  in  the 
perpendicular  through  a,  the  closing  line  of  the  polygon  o  d 
would  have  been  horizontal.  It  is,  however,  indifferent  at 
what  inclination  a*  d  may  lie,  but  we  may  if  we  wish  make  it 
horizontal  now,  and  then  lay  off  from  its  new  intersections  with 
P!  and  P2  along  the  directions  of  these  forces,  the  ordinates 
already  found  at  £  and  c,  and  join  the  points  thus  obtained  with 
the  ends  of  o  d  (i.e.,  with  its  intersections  with  the  verticals 
through  the  supports).  The  ordinates  of  the  new  polygon 
thus  found  will  be  for  any  point  the  same  as  before,  and  will 
also  be  perpendicular  to  the  beam. 

[NOTE. — Had  we  taken  the  forces  precisely  as  above  but  in  reverse  order, 
the  force  line  would  be  reversed,  and  we  should  have  0  and  2  in  place  of  1, 
and  1  in  place  of  0  and  2 ;  that  is,  in  place  of  O  1  we  should  have  O  0  and 
C  2.  Constructing  then  the  equilibrium  polygon  by  drawing  a  line  paral- 
lel to  new  O  0  to  intersection  with  new  Pi,  then  parallel  to  new  C  1  to  in- 
tersection with  new  P2,  then  parallel  with  new  O  2  to  intersection  with 
vertical  through  B,  and  finally  joining  this  last  point  with  intersection  of 
the  first  line  drawn  (C  0)  with  vertical  through  A,  we  have  at  first  sight  a 
very  different  equilibrium  polygon.  This  new  polygon  will  consist  of  two 
parts.  If  the  ordinates  in  one  of  these  parts  are  considered  positive,  those 
iii  the  other  must  be  negative.  The  difference  of  the  ordinates  in  these  two 
portions  for  any  point,  will  give  the  same  result  as  above.  This,  by  mak- 
ing the  above  construction,  the  reader  can  easily  prove.] 


CHAP.  V.]  MOMENT  OF  RUPTURE  OF  PARALLEL  FORCES.         41 

41.  Many  other  problems  will  readily  occur,  which  may  in  a 
similar  manner  be  solved.  The  weights  may  have  any  position, 
number  and  intensities  desired  ;  in  any  and  every  case  we  have 
only  to  construct  with  assumed  pole  distance  the  corresponding 
equilibrium  polygon,  and  we  obtain  at  once  the  moments  at 
every  point.  By  the  use  of  convenient  scales,  numerical  results 
may  be  obtained  which  may  be  checked  by  calculation,  and 
the  practical  value  and  accuracy  of  the  method  thus  demon- 
strated. 

The  above  principles  will  be  sufficient  for  the  solution  of  any 
such  problem  wrhich  may  arise,  and  we  shall  therefore  content 
ourselves  with  the  above  general  indication  of  the  method  of 
procedure,  and  pass  on  to  the  consideration  of  a  few  cases 
where  the  above  needs  slight  modification,  and  which,  from 
their  practical  importance,  and  the  ease  with  which  they  may 
be  treated  graphically,  seem  worthy  of  special  notice. 

1ST.   BEAM   OR   AXLE — LOAD   INCLINED   TO  AXIS.*       [Fig.  25,  PL  6.] 

We  have  here  simply  to  draw  the  "  closing  line  "AC  paral- 
lel to  the  beam  or  axle.  From  d  draw  d  B  parallel  to  the  force 
P,  then  draw  A  B  in  any  direction  at.  pleasure,  and  join  B  C. 
We  have  thus  the  equilibrium  polygon  ABC,  the  ordinates  to 
which,  as  d  B,  parallel  to  the  force  P,  will  give  the  moments, 
provided  we  know  the  corresponding  pole  distance. 

But  this  can  easily  be  found.  As  we  have  already  seen,  the 
force  polygon  being  given,  the  equilibrium  polygon  may  be 
easily  constructed.  Inversely,  the  equilibrium  polygon  being 
given,  the  force  polygon  may  be  constructed.  Thus  from  A 
draw  A  c  equal  and  parallel  to  P,  and  then  draw  c  C^  parallel 
to  B  C.  A  a  and  J  c  are  the  vertical  reactions  P!  and  P2 ;  a  b 
is  the  horizontal  component  of  the  force  which  must  be  resisted 
at  one  or  both  of  the  ends ;  and  the  moments  at  any  point  are 
given  by  the  ordinates  parallel  to  P  multiplied  by  the  perpen- 
dicular distance  from  CL  to  A  c.  If  we  suppose  the  force  P,  as 
in  the  Fig.,  as  causing  two  opposite  vertical  forces,  instead  of 
acting  directly  upon  the  axis,  we  have  only  to  prolong  A  B  to 
B!  and  join  B!  62,  and  then  the  ordinates  of  A  B!  B2  C  parallel 
to  P  or  A  c,  multiplied  by  H  (perpendicular  distance  from  Ct 
to  A  c)  will  give  the  moments. 

*  See  Der  Constructeur,  Reuleaux. 


4-2        MOMENT  OF  RUPTURE  OF  PARALLEL  FORCES.  [CHAP.  V. 
2D.  FORCE  PARALLEL  TO  AXIS.   [Fig.  26,  PL  6.] 

We  have  an  example  of  this  case  in  the  "  bayonet  slide  "  of 
the  locomotive  engine. 

We  have  here  two  pairs  of  forces,  the  reactions  Vt  and  V2 
and  the  forces  over  B!  and  B2.  The  points  of  application  of 
these  last  change  of  course  periodically,  but  for  any  assumed 
position  the  moments  are  easily  found.  Thus  draw  A  B!  at 
pleasure,  and  C  B2  parallel  to  it,  and  join  Bt  B2  arid  A  C,  and 
we  have  at  once  the  equilibrium  polygon.  To  find  the  corre- 
sponding force  polygon^  suppose  Px  applied  at  5,  and  join  ~b  with 
the  other  support.  Make  b  c  equal  to  P  then  c  d  =  V2.  Lay 
off  then  A  a  =  c  d  —  V2  and  draw  a  Cl5  which  is  the  pole  dis- 
tance. Draw  GI  e  parallel  to  B!  B2.  Then  A  e  and  e  A  are  the 
forces  acting  over  Bg  and  B1?  and  A  a  is  the  reaction  V^  The 
case  is,  indeed,  precisely  similar  to  that  in  Art.  40. 

[NOTE. — The  moment  area  should  properly  be  turned  over  upon  A  C  as 
an  axis,  so  that  A  a  should  be  laid  off  and  e  fall  lelow  A.  This  can,  how- 
ever, cause  no  confusion.] 

The  application  of  the  method  to  car  axles,*  crane  standards, 
and  a  large  number  of  similar  practical  cases  in  Mechanics  is 
obvious.  The  formulae  for  many  of  these  cases  are  too  com- 
plex  for  practical  use  ;  in  some,  no  attempt  at  investigation  of 
strain  is  ever  made,  the  proportions  being  regulated  simply  by 
"  Engineering  precedent "  or  rules  of  thumb.  Those  familiar 
with  the  analytical  discussion  of  such  cases  will  readily  recog- 
nize the  great  practical  advantages  of  the  Graphical  Method. 

3D.     BEAM   OR   AXLE   ACTED   UPON   BY   FORCES   LYING   IN  DIFFERENT 

PLANES. 

The  analytical  calculation  in  such  a  case  for  instance  is  of 
considerable  intricacy,  but  by  the  graphical  method,  on  the 
contrary,  the  difficulty  of  investigation  is  scarcely  greater  than 
before. 

Thus,  let  Fig.  27,  PL  7,  represent  a  beam  acted  upon  by  two 
forces  P!  and  P2  not  in  the  same  plane. 

First,  we  draw  the  force  polygons  A  C^  M  and  D  O2  2  for  the 
forces  P!  and  P2,  having  both  the  same  pole  distance  G  Ol  = 
O2  H,  the  pole  O2  being  so  taken  that  the  closing  lines  of  the 

*  Der  Constructeur,  Reuleaux,  pp.  215-222. 


CHAP.  V.J  MOMENT  OF  RUPTURE  OF  PARALLEL  FORCES.         43 

corresponding  polygons  A  V  D  and  A  c"  D  coincide.  This  is 
easily  done,  as  if  the  closing  line  of  the  second  polygon  for  any 
assumed  position  of  O2  (O2  H  being  equal  to  G  OJ  does  not  co- 
incide with  A  D,  the  ordinate  at  c'  can  be  laid  off  from  C  and 
A  G"  D  thus  found  in  proper  position,  and  then  the  pole  O2  can 
be  located.  It  will  evidently  be  at  the  intersection  of  the  ver- 
tical O2  O'2  with  G"  D. 

The  two  force  polygons  being  thus  formed,  we  construct  the 
polygon  A  C"  D  by  drawing  lines  B  B",  E  E",  C  C",  etc.,  so 
that  their  angles  with  the  vertical  shall  be  equal  to  the  angle 
between  the  planes  of  the  forces,  and  making  them  equal  to 
the  ordinates  B  5",  E  e'1 ',  C  c",  etc.,  respectively.  Join  V  B", 
e'  E",  /'  F",  G'  C",  etc.,  "and  lay  off  the  ordinates  B  b,  E  e,  F/, 
C  c,  etc.,  respectively  equal.  The  ordinates  to  the  polygon 
thus  obtained,  viz. :  A  5  efc  D  multiplied  by  the  pole  distance 
Oi  G  or  O2  H,  give  the  moments  at  any  point.  A  1)  and  G  D 
are  straight  lines,  ~b  efc\&  a  curve  (hyperbola).  If  we  drop 
verticals  through  C^  and  O2,  and  draw  the  perpendiculars  O/  M, 
O'2K ;  A  M  is  the  reaction  R1;  and  D  K  the  reaction  R2,  both 
measured  to  the  scale  of  the  force  polygon.  Their  directions 
are  found  by  the  composition  of  A  G  and  H  2  and  D  H  and 
G  M  respectively,  under  the  angle  of  the  forces. 

4TH.     COMBINED    TWISTING   AND   BENDING   MOMENTS. 

In  many  constructions  pieces  occur  which  are  subjected  at 
the  same  time  to  both  bending  and  twisting  moments.  Both 
can  be  represented  and  given  by  moment  areas.  Thus.,  Fig. 
28,  PL  7,  represents  an  axle  turning  upon  supports  at  A  and  B 
and  having  at  C  a  wheel  upon  which  the  force  P  acts  tangenti- 
ally.  We  have  then  a  moment  of  torsion  Mt  =  P  R  and  reac- 

*  o  ft 

tions  Pt  —  P  -    -  and  P2  =  P  —    -  ;  s  being  the  distance  of  P 

a  i  S  a  -\~  S 

from  B,  and  a  of  P  from  A. 

Let  the  bending  moments  be  represented  by  the  ordinates  to 
the  polygon  a  C  ~b ;  then  laying  off  a  o  equal  to  P  and  drawing 
o  O  parallel  to  ~b  c,  we  find  the  corresponding  pole  distance 
O  &,  and  the  reactions  P!  and  P2  equal  to  Jc  a  and  o  Jc  respec- 
tively. 

Now,  in  the  force  polygon  O  a  o  thus  found,  at  a  distance 
from  O  equal  to  R,  draw  a  line  m  n  parallel  to  P.  This  line 
m  n  evidently  gives  for  the  same  pole  distance  the  moment  of 


44  MOMENT   OF   RUPTURE    OF   PARALLEL   FORCES.      [CHAP.  V. 

torsion  P  x  R.  Laying  off  C'C^  =  b  &',  —  in  n,  we  have  the 
torsion  rectangle  Cx  b'  b  C'. 

.  Now  the  combined  moment  of  torsion  Mt  and  bending  Mb  is 
|  Mb-f  f  v/Mg  +  Mp.  We  make  then  C'  C0  equal  to  f  C'  Ct  = 
£  m  n  and  C  C2  equal  to  f  C  C'  =  Mb,  and  draw  C2  b.  Then 
any  segment  of  any  ordinate,  as/yj,  is  -f  offf.  Revolve  now 
C'  C0  with  C'  as  a  centre,  round  to  C'0  and  join  C'0  C2.  Then 
C2  C'0  is  equal  to  f  V^b  H-M?>  and  therefore  with  C2  as  centre 
revolving  C2  C'0  to  C3,  we  find  the  point  C8,  C  C3  being  equal  to 
f  Mb  +  f  \/M^  4-  MjL  In  the  same  way  we  find  any  other 
point  as/3,  by  laying  off//',,  equal  to//0,  joining/3  and/'0  and 
making  f2  fs  equal  to  f%  fQ.  The  line  C3  f$  b0  thus  found  is  a 
hyperbola,  and  the  ordinates  between'  it  and  b  C  give  the  com- 
bined moments  [for  pole  distance  O  Jc]  at  any  point. 

[NOTE. — We  suppose  the  axle  to  turn  freely  at  A,  and  the  working  point 
or  resistance  beyond  B ;  hence  the  moments  left  of  the  wheel  are  given  by 
the  ordinates  to  a  O.] 

/ 

5TH.    APPLICATION   TO    CRANK   AND    AXLE. 

The  above  finds  special  and  important  application  in  the  case 
of  the  crank  and  axle. 

Thus  in  PL  8,  Fig.  29,  let  E  D  C  B  be  the  centre  line  of  crank 
and  shaft.  Lay  off  a  P  equal  to  the  force  P  acting  at  A,  choose 
a  pole  o  and  draw  o  a  o  P  and  the  parallels  o  a  and  a  E.  Join 
E  and  d  and  draw  o  Pj  parallel  to  E  d.  Then  P  P!  is  the 
downward  force  at  E  and  P1  a  the  upward  reaction  at  D.  The 
ordinates  to  E  d  a  to  pole  distance  o  P,  give  the  bending  mo- 
ments for  the  shaft.  Make  a  F  equal  to  the  lever  arm  R,  then 
F  G  is  the  moment  P  R,  and  we  unite  this  as  above  with  the 
bending  moments  and  thus  find  the  curves  c'  d'  e'  the  ordinates 
to  which  give  the  combined  moments  at  every  point  of  the 
shaft  [see  4th]. 

For  the  arm  B  C,  make  the  angle  aQ  B  C  equal  to  D  a  d, 
and  then  the  horizontal  ordinates  to  aQ  B  give  the  bending  mo- 
ments for  the  arm.  Make  C  c0  equal  to  C  G  and  we  have  the 
torsion  rectangle  C  CQ  bQ  B,  and  as  in  the  previous  case  we  unite 
the  two  and  thus  find  the  curve  b0  h  F,  the  horizontal  ordinates 
to  which  from  B  C  give  the  required  combined  moments,  to 

*  Der  Constructeur,  Reuleaux,  p.  52,  Art.  18. 


CHAP.  V.]   MOMENT  OF  RUPTURE  OF  PARALLEL  FORCES.        45 

polo  distance  o  P.  Thus  h'h^  =  f  H  A0j  H  i  —  f  B  50,  and 
H  h  =  h,  h'  +  ti  i  =  f  Mb  +  |  /MS  +  M?. 

The  application  of  the  method  when  the  crank  is  not  at  right 
angles  to  the  shaft,  as  also  when  the»crank  is  double,  and  gener- 
ally in  the  most  complicated  cases,  is  equally  simple  and  satis- 
factory. Our  space  forbids  any  more  extended  notice  of  these 
applications,  and  we  must  refer  the  reader  to  Der  Constructeiir, 
by  F.  Reuleaux,  Braunschweig,  1872,  for  further  illustrations 
and  applications  of  the  method  to  the  solution  of  various  practi- 
cal mechanical  problems. 

42.  Continuous  Loading— Load  Area. — Thus  far  we  have 
considered  only  concentrated  loads.  But  whatever  may  be  the 
law  of  load  distribution,  if  this  law  is  known,  we  can  represent 
it  graphically  by  laying  off  ordinates  at  every  point,  equal  by 
scale  to  the  load  at  that  point.  We  thus  obtain  an  area  bounded 
by  a  broken  line,  or  for  continuous  loading,  by  a  curve,  the 
ordinates  to  which  give  the  load  at  any  point.  This  load  area 
we  can  divide  into  portions  so  small  that  the  entire  area  may 
be  considered  as  composed  of  the  small  trapezoids  thus  formed. 
If,  for  instance,  we  divide  the  load  area  into  a  number  of  trape- 
zoids of  equal  width,  as  one  foot  one  yard,  etc.,  as  the  case  may 
be,  then  the  load  upon  each  foot  or  yard  will  be  given  by  the 
area  of  each  of  these  trapezoids.  If  the  trapezoids  are  suffi- 
ciently numerous,  we  may  consider  each  as  a  rectangle  whose 
base  is  one  foot  or  one  yard,  etc.,  as  the  case  may  be,  and  whose 
height  is  the  mean  or  centre  height.  The  weight  therefore  for 
each  trapezoid  acts  along  its  centre  line.  We  thus  obtain  a 
system  of  parallel  forces,  each  force  being  proportional  to  the 
area  of  its  corresponding  trapezoid,  and  equal  by  scale  to  the 
mean  height  or  some  convenient  aliquot  part  of  this  height. 
We  can  then  form  the  force  polygon  /  choose  a  pole ;  draw 
lines  from  the  pole  to  the  forces ;  and  then  parallels  to  these 
lines,  thus  forming  the  string  or  equilibrium  polygon;  and  so 
obtain  the  graphical  representation  of  the  moments  at  every 
point. 

Since,  however,  the  polygon  in  this  case  approximates  to  a 
curve,  that  is,  is  composed  of  a  great  number  of  short  lines,  the 
above  method  is  subject  to  considerable*  inaccuracy,  as  errors 
multiply  in  going  along  the  polygon. 

This  difficulty  can,  however,  be  easily  overcome. 

Thus  we  may  divide  the  load  area  into  two  portions  only,  and 


46  MOMENT   OF   RUPTURE    OF   PARALLEL   FORCES.      [CHAP.  V. 

then  draw  the  force  and  equilibrium  polygon,  considering  each 
portion  to  act  at  its  centre  of  gravity,  and  so  obtain  an  equili- 
brium polygon  composed  of  three  lines  only.  These  lines  will 
be  tangents  to  the  equilibrium  curve.  (Art.  76.)  We  thus  have 
three  points  of  the  curve,  and  its  direction  at  these  points. 
In  this  manner  we  may  determine  as  many  points  as  may  be 
necessary,  without  having  the  sides  of  the  polygon  so  short  or 
so  numerous  as  to  give  rise  to  inaccuracy. 

43. — The  above  will  appear  more  plainly  by  consideration 
of  a 

BEAM  UNIFORMLY  LOADED. 

The  curve  of  load  distribution  becomes  in  this  case  a  straight 
line.  The  load  area  is  then  a  rectangle,  and  hence  the  load  per 
unit  of  length  is  constant.  Let  us  now  divide  this  load  area 
[Fig.  30,  PI.  8],  into  four  equal  parts,  and  considering  each  por- 
tion as  acting  at  its  centre  of  gravity,  assume  a  scale  of  force, 
and  draw  the  force  polygon.  Since  in  this  case  the  reactions  at 
the  supports  must  be  equal,  we  take  the  pole  C,  in  a  perpendi- 
cular to  the  force  polygon  at  the  middle  point.  This  causes  the 
closing  line  of  the  equilibrium  polygon  to  be  parallel  to  the 
beam  itself,  which  is  often  convenient.  We  now  draw  C  0,  C  1, 
etc.,  and  then  form  the  polygon  0  a  c  e  g  h.  The  lines  0  a,  a  c,  • 
c  e,  etc.,  of  this  polygon,  are  tangent  to  the  moment  curve  at 
the  points  5,  d,f,  0  and  A,  where  the  lines  of  division  prolonged 
meet  the  sides.  The  curve  can  now  be  easily  constructed,  as 
will  appear  from  the  next  Art. 

44.  moment  Curve  a  Parabola. — Suppose  we  had  divided 
the  load  area  into  only  two  parts,  of  the  length  x  and  I— x  [Fig. 
30,  PL  8].  Then  the  moment  polygon  would  be  o  a  k  h,  and 
the  horizontal  projection  of  the  tangent  a  Jc  would  be  %  x  4-  i 
(t±xj-ll. 

That  is,  the  horizontal  projection  of  any  tangent  to  the  mo- 
ment curve  is  constant.  But  this  is  a  property  of  the  parabola. 
The  moment  curve  for  a  uniform  load  is  therefore  a  parabola, 
symmetrical  with  respect  to  the  vertical  through  the  centre  of 
the  l>eam. 

If,  then,  we  divide  o  C  and  C  h  into  equal  parts,  and  join  cor- 
responding divisions  above  and  below,  we  can  construct  any 
number  of  tangents  in  any  position. 

["NOTE. — We  may  prove  analytically  that  the  moment  curve  is  a  parabola, 


CHAP.  V.]  MOMENT  OF  RUPTURE  OF  PARALLEL  FORCES.         47 

and  hence  that  the  line  a  Tc  must  le  a  tangent.     Thus  the  moment  at  any 
point  is 

p  being  the  load  per  unit  of  length,  I  the  length,  and  the  reaction  at  sup- 

f)  1  f) 

port  therefore  ~-.     Hence  y  =  — —  (Ix  —  x1}  for  origin  0. 

a  a  H 

When  the  origin  is  at  d,  representing  horizontal  distances  by  y'  and  ver- 
tical by  x1,  we  have  x  =  -  —  y',  and  y  =  h  —  #',  k  being  the  ordinate  at 

9 

^   72 

middle  = 

Hence  by  substitution 


or  reducing 

2H 

f  =  T  x 

which  is  the  equation  of  a  parabola  having  its  vertex  at  <#.] 

We  may  of  course  take  the  pole  anywhere,  and  hence  H  may 
have  any  value.  It  is  in  general  advantageous  in  such  cases 

(i.e.,  for  uniform  load)  to  take  H  =  *—•    We  have  then 

f  =  I  so, 
and  for  y  —  •%,  or  for  the  middle  ordinate.  we  have  x  =  j* 

To  draw  the  moment  curve  we  have  then  simply  to  lay  off 
the  middle  ordinate  equal  to  Jth  the  span.  The  curve  can  then 
be  constructed  in  the  customary  way  for  a  parabola.  Any 

pi 
ordkiate  to  this  curve  multiplied  by  H  =  ^—  will  then  give  the 

moment  at  that  point. 

Enough  has  probably  now  been  said  to  illustrate  the  applica- 
tion of  our  method  to  the  determination  of  the  moment  of  rota- 
tion, bending  moment,  or  moment  of  rupture.  The  reader 
will  have  no  difficulty  in  applying  the  above  principles  to  any 
practical  case  that  may  occur. 

It  will  be  observed  that  the  customary  curve  of  moments  in 
the  graphic  methods  at  present  in  general  use,  comes  out  as 
a  particular  case  of  the  equilibrium  polygon  for  uniform 
load. 

This  polygon  has  other  interesting  properties,  which  we  shall 
notice  hereafter.  For  instance,  just  as  its  ordinates  [Fig.  30] 
are  proportional  to  the  bending  moments  or  moment  of  rotation, 


4:8        MOMENT  OF  RUPTURE  OF  PARALLEL  FORCES.   [dlAP.  V. 

so  also  its  area  is  proportional  to  the  moments  of  the  moments, 
or  the  moment  of  inertia  of  the  load  area. 

As  to  the  shearing  force  at  any  point  of  a  beam  submitted 
to  the  action  of  parallel  forces,  the  reactions  at  the  ends  being 
easily  found  as  above  by  a  line  parallel  to  the  closing  line  in 
the  force  polygon,  we  have  only  to  remember  that  the  shear  at 
any  point  is  equal  to  the  reaction  at  one  end,  minus  all  the 
weights  between  that  end  and  the  point  in  question. 

Thus  for  a  uniformly  distributed  load  we  have  simply  to  lay 
off  the  reactions  which  are  equal  to  one-half  the  load,  above 
and  below  the  ends,  and  draw  a  straight  line,  which  thus  passes 
through  the  centre  of  the  span.  The  ordinates  to  this  line  are 
evidently  then  the  shearing  forces.  If  we  have  a  series  of  con- 
centrated loads,  we  have  a  broken  line  similar  to  A\  1'  1"  2', 
etc.,  Fig.  32,  PI  7,  where  each  successive  weight  as  we  arrive  at 
it,  is  subtracted  from  the  preceding  shear. 

44.  Beam  continuously  Loaded  and  also  Subjected  to 
the  Action  of  Concentrated  Loads. — In  practice  we  have 
to  consider  not  only  a  continuously  distributed  load,  such  as  the 
weight  of  the  truss  or  beam  itself,  but  also  concentrated  forces, 
such  as  the  weight  of  cars,  locomotives,  etc.,  standing  upon  or 
passing  over  the  truss. 

In  PL  8,  Fig.  31,  we  have  a  continuous  loading  represented 
by  the  load  area  A  a  b  B,  and  in  addition  four  forces  P'w. 
Now,  since  the  total  moment  about  any  point  is  equal  to  the 
sum  of  the  several  moments,  we  can  treat  each  method  of  load- 
ing separately  and  then  combine  the  results.  Thus  with  the 
force  polygon  (b)  we  obtain  the  equilibrium  polygon  A'  1  2  3 

B'  for  the  continuous  loading,  and  with  the  force  polygon 

(a)  the  equilibrium  polygon  A'  I"  2"  3" B"  for  the  con- 
centrated loads.  If  now  in  (b)  we  draw  C  L  parallel  to  the 
closing  line  A'  B',  and  in  (a)  C"  L'  parallel  to  the  closing  line 
A'  B",  we  obtain  at  once  the  reactions  at  the  supports  for  each 
case. 

Thus  for  continuous  loading  we  have  L  0  for  reaction  at  A, 
and  10  L  for  reaction  at  B ;  for  the  concentrated  loads,  L'  0' 
at  A  and  4'  L'  at  B.  These  reactions  hold  the  beam  in  equi- 
librium. 

For  any  cross-section  ^,  the  shear  to  the  right  is  composed  of 
the  two  components  L  7  and  I/  3'  (i.e.,  is  equal  to  the  reactions 
minus  the  forces  between  cross-section  and  support).  The  mo- 


CHAP.  V.]  MOMENT  OF  KUPTUKE  OF  PARALLEL  FORCES.         49 

ment  of  L  7  is  given  by  the  ordinate  o  y  to  the  corresponding 
polygon,  and  we  may  consider  L  7  as  acting  at  the  point  of 
intersection  a  of  the  side  7  8  with  A!  B'  (Art.  38).  In  the  same 
way  L'  3'  acts  at  b.  We  may  unite  both  these  reactions  and 
find  the  point  of  application  of  their  resultant  £,  by  laying  off 
in  force  polygon  (b)  7  b  equal  to  I/  3',  and  then  constructing 
the  corresponding  equilibrium  polygon  e  a  d  c.  The  resultant 
R  passes  through  c.  This  construction  remains  the  same  evi- 
dently, even  when  the  points  a  and  b  fall  at  different  ends  of 
the  beam,  as  may  indeed  happen.  The  components  will  then 
have  opposite  directions,  and  must  be  subtracted  in  order  to 
obtain  the  resultant. 

The  total  moment  of  rotation  at  y  is  proportional  to  the  sum 
of  m  n  and  o  y.  The  greatest  strain  is  where  this  sum  is  a 
maximum.  In  order  to  perform  this  summation  and  ascertain 
this  point  of  maximum  moment  it  is  advantageous  to  construct 
another  polygon  instead  of  A'  1"  2",  etc.,  whose  closing  line 
shall  coincide  with  A'  B'.  This  is  easy  to  do,  by  drawing  in 
force  polygon  (a),  I/  C'  parallel  to  A'  B',  and  taking  a  new  pole 
C'  the  same  distance  out  as  before,  that  is,  keeping  H  constant, 
and  then  constructing  the  corresponding  polygon  A.'  V  2'  3',  etc. 

Thus  the  ordinate  p  y  gives  the  total  moment  at  y.  We  can 
make  use  here  also  of  the  principle  that  the  corresponding  sides 
of  the  two  polygons  must  intersect  upon  the  vertical  through 
A'  (Art.  26).  We  have  thus  the  total  moment  at  any  point,  and 
can  easily  determine  the  point  of  maximum  moment  or  cross- 
section  of  rupture.  This  point  must  necessarily  lie  between 
the  points  of  maximum  moments  for  the  two  cases,  or  coincide 
with  one  of  them.  In  the  Fig.  this  point  coincides  with  the 
point  of  application  of  P'2. 

45.  Case  of  Uniform  Load. — If  the  continuous  load  is  uni- 
formly distributed  we  can  obtain  the  above  result  without 
being  obliged  to  draw  the  curve.  As  in  this  case  we  have  a 
very  short  construction  for  the  determination  of  the  point  of 
greatest  moment,  it  may  be  well  here  briefly  to  notice  it. 

If  we  erect  ordinates  along  the  length  of  the  beam  as  an  axis 
of  abscissas,  equal  to  the  sum  of  the  forces  acting  beyond  any 
cross-section,  the  line  joining  the  end  points  of  these  ordinates 
has  a  greater  or  less  inclination  to  the  axis  according  as  the 
uniform  load  is  greater  or  smaller.  At  the  points  of  applica- 
tion of  the  concentrated  loads  this  line  is  evidently  shifted 
4 


50  MOMENT   OF   RUPTTJKE    OF   PARALLEL   FOKCES.      [CHAP.  V. 

parallel  to  itself.  Since  at  the  point  of  maximum  strain  the 
sum  of  the  forces  either  side  is  zero,  this  point  is  given  by  the 
intersection  of  the  broken  line  thus  found  with  the  axis. 

Thus  in  PL  7,  Fig.  32,  let  A  B  be  the  beam  sustaining  a  uni- 
form load,  and  also  the  concentrated  loads  Pt  P2  P3  P4.  The 
reaction  of  the  uniform  load  at  the  supports  is  equal  to  half 
that  load.  To  find  the  reactions  for  the  concentrated  loads  we 
draw  the  force  polygon  01234,  choose  a  pole  C,  then  con- 
struct the  equilibrium  polygon  A'  1  2  3  4  B',  and  parallel  to 
A'  B'  draw  C  L.  L  0  and  L  4  are  the  reactions  at  A  and  B. 
ISTow  through  L  draw  A0  L  horizontal,  make  it  equal  to  the 
length  of  the  beam,  and  take  it  as  axis  of  abscissas.  [It  is  of 
course  advantageous  here  to  lay  off  the  forces  along  the  verti- 
cal through  B,  as  done  in  the  Fig.  Then  A0  falls  in  the  vertical 
through  A  and  10  20  30  40  are  directly  under  the  forces  them- 
selves.] 

The  ordinate  to  be  laid  off  at  A0  is  equal  to  L  0  +  half  the 
uniform  load.  Between  A0  and  10  the  line  A'x  1'  is  inclined  to 
the  axis  at  an.  angle  depending  upon  the  uniform  load.  Lay  off 
L  U  equal  to  this  load  and  draw  A0  U.  A\  1'  must  be  parallel 
to  this  line.  At  1'  the  line  A.\  I'  is  shifted  to  1",  so  that  1  1 " 
is  the  load  Pt.  Then  1"  2'  is  parallel  as  before  to  A0  U,  and 
2'  2"  is  the  load  P2,  and  so  on.  The  intersection  20  with  A0 1* 
gives  the  point  of  maximum  moment  or  cross-section  of  rup- 
ture. The  force  P2  at  this  point  in  our  Fig.  is  divided,  as  shown 
by  L  in  the  force  polygon,  into  two  portions,  one  of  which  is  to 
be  added  to  the  forces  left,  the  other  to  the  forces  right.  The 
ordinate  y0  y'  at  any  point  gives  the  shear  or  sum  of  the  forces 
acting  at  that  point.  This  force  acts  up  or  down  according  as 
the  ordinate  is  above  or  below  the  axis. 

Moreover,  the  area  between  the  broken  line  and  axis  A0  L, 
limited  by  this  ordinate,  gives  the  moment  of  rotation  of  the 
forces  beyond  the  section  y,  areas  below  the  axis  being  nega- 
tive. For  a  section  at  2,  therefore,  we  have  area  A0  A\  1'  1" 
2'  20,  minus  20  2"  3'  3"  z'  20,  or  what  is  the  same  thing,  the  area 
ZQ  z'  4'  4"  B\  L,  since  the  sum  of  the  moments  of  all  the  forces 
is  zero. 

46.  Influence  of  a  Concentrated  Load,  pa§§iiig  over  the 
Beam. — If  in  addition  to  the  already  existing  uniform  and 
concentrated  loads,  a  new  force  operates,  we  have  by  (44)  simply 
to  construct  for  this  new  force  its  force  and  equilibrium  poly- 


CHAP.  V.]  MOMENT  OF  RUPTURE  OF  PARALLEL  FORCES.         51 

gon,  and  unite  the  forces  and  moments  thus  found  with  those 
already  existing. 

In  PL  7,  Fig.  32,  we  have  assumed  a  new  force  P\  near  the 
left  support.  The  force  polygon  is  0'  1'  C',  the  pole  distance 
being  taken  the  same  as  before.  For  any  one  position  of  this 
force  we  have  then  the  equilibrium  polygon  A'  1'  B;/,  and 
drawing  a  parallel  C'  L'  to  A'  B"  we  obtain  the  reactions  0'  I/ 
and  L'  1',  which  must  be  added  to  the  reactions  already  ob- 
tained. 

If  now  we  take  a  section  y  between  P\  and  the  point  of  max- 
imum moment  20  before  found,  the  sum  of  the  forces  either 
side  of  this  section  undergoes  the  following  changes :  Upon 
the  side  where  P^  lies,  and  the  point  20  does  not  lie,  where 
therefore  the  sum  was  originally  an  upward  force,  we  have  the 
downward  force  I/  V  (equal  to  algebraic  sum  L'  0'  4-  0'  1'). 
The  sum  of  the  forces  at  the  section,  or  the  shearing  force,  is 
therefore  diminished. 

The  total  rotation  moment  is,  however,  increased  by  the 
amount  indicated  by  m  n.  Both  changes,  that  of  the  sum  of 
the  forces  and  the  moment  'of  rotation,  increase  as  P\  ap- 
proaches y,  and  are  therefore  greatest  when  P\  reaches  y. 

If  P\  passes  ?/,  this  point  is  in  the  same  condition  as  z  with 
reference  to  the  former  position  of  P\ ;  that  is,  the  force  and 
point  20  are  now  both  on  the  same  side  of  the  section.  For  z, 
then,  the  original  downward  force  to  the  left  is  increased  by  the 
force  I/  V.  To  the  right  the  upward  force  is  increased  by  1'  L'. 
In  like  manner  the  moment  of  the  forces  beyond  z  is  increased 
by  the  amount  indicated  by  op.  This  change  is  greatest  when 
P\  reaches  z. 

Therefore  when  a  load  passes  over  the  beam  the  sum  of  the 
shearing  forces  is  diminished  in  all  sections  between  it  and  the 
original  point  of  greatest  moment,  and  increased  m  sections  be- 
yond this  point,  while  the  moment  of  rotation,  or  bending 
moment,  for  all  cross-sections  is  increased.  These  changes 
moreover  increase  for  any  section  as  the  load  approaches  that 
section.  The  shear  at  any  point  is  therefore  least,  and  the  mo- 
ment greatest,  when  the  load  reaches  that  point.  As  soon,  how- 
ever, as  the  load  passes  this  point,  the  shear  passes  suddenly 
from  its  smallest  to  its  greatest  opposite  value,  and  then  dimin- 
ishes as  the  load  recedes,  together  with  t'he  moment  of  rotation. 
On  the  other  side  of  the  point  20  of  original  greatest  moment, 


52  MOMENT   OF   RUPTURE    OF  PARALLEL   FORCES.      [CHAP.  V. 

the  shear  and  moment  increase  as  the  load  approaches,  and 
become  greatest  for  any  point  when  the  load  reaches  that  point. 
At  the  moment  of  passing,  these  greatest  values  pass  to  their 
smallest  values,  and  increase  afterwards,  as  the  load  recedes. 

Since  by  the  introduction  of  the  load  the  shear  for  points 
upon  one  side  of  20  is  diminished  (between  20  and  the  load),  and 
on  the  other  side  increased,  and  the  greatest  moment  is  at  the 
point  where  the  shear  is  zero,  it  follows  that  the  point  of  greatest 
moment  moves  in  general  towards  the  load.  At  a  certain  point, 
then,  both  meet.  As  the  load  then  advances  this  point  accom- 
panies it,  passes  with  it  the  original  position,  and  follows  it  up 
to  the  point  where  it  would  have  met  the  same  load  coming  on 
from  the  other  side.  From  this  point,  as  the  load  continues  to 
recede,  it  returns,  and  finally  reaches  its  original  position  as  the 
load  arrives  at  the  further  end. 

It  is  evidently  of  interest  to  learn  the  position  of  these  two 
points,  where  the  load  meets  and  leaves  the  point  of  greatest 
moment,  or  cross-section  of  rupture,  and  this  in  Fig.  32  we  can 
easily  do. 

When  P\  arrives  at  1',  we  have  evidently  the  reactions  by 
laying  off  L  E  equal  to  P'1?  drawing  A0  E,  and  through  its 
intersection  with  the  vertical  through  the  weight  drawing  the 
horizontal  A'0  B'0.  L  B'0  is  then  the  increase  of  reaction  at  B  due 
to  P\.  The  entire  reaction  is  B'0  B'1?  and  the  broken  line  A\ 
1'  1",  etc.,  holds  good  still,  if  we  merely  change  the  axis  from 
AO  L  to  A'0  B'0.  The  point  of  greatest  moment,  which  is  still 
the  intersection  of  the  broken  line  with  the  new  axis,  in  the 
present  case  is  not  changed  by  reason  of  the  overpowering  in- 
fluence of  P2.  It  does  not  move  to  meet  the  load,  but  awaits  it 
until  it  reaches  P2,  and  until,  therefore,  the  new  axis  takes  the 
position  A"0  B"0. 

If,  however,  the  force  P\  comes  on  from  the  right,  we  have 
the  reactions  for  any  position  as  z,  by  laying  off  A0  E'  equal  to 
•P'u  drawing  L  E',  and  then  the  horizontal  A'"0  B'"0  through  the 
intersection  of  L  E',  with  the  vertical  through  z.  Then  A0  A'"0 
is  the  reaction  at  A,  due  to  this  position  of  the  load.  The  in- 
tersection #',  corresponding  to  a?,  shows  the  point  to  which  the 
point  of  greatest  moment  20  moves  to  meet  the  load.  As  the 
load  passes  towards  the  left,  this  point  moves  towards  the  right, 
and  both  come  together  evidently  at  the  point  V^  correspond- 
ing to  the  new  axis  A0iv  B0iv.  The  point  of  greatest  moments 


CHAP.  V.]     MOMENT   OF   RUPTURE   OF   PARALLEL   FORCES.  53 

passes  then  from  20  to  V0,  and  beyond  these  two  limits  it  can 
never  pass. 

Our  construction,  then,  is  simply  to  lay  off  the  load  in  oppo- 
site directions  perpendicularly  from  each  end  of  the  axis  A0  L, 
and  join  the  end  points  A0  E  and  L  E'.  The  intersections  of 
these  lines  with  the  diagram  of  shear  give  the  points  20  and 
V0  required. 

47.  Load  Systems.* — Concentrated  loads  occur  in  general 
in  practice  in  a  certain  succession,  as  for  instance  the  forces 
acting  at  the  points  of  contact  of  the  wheels  of  a  train  of  cars 
passing  over  the  beam,  and  it  is  necessary  then  to  investigate 
the  influence  of  different  positions  of  the  train.  It  evidently 
amounts  to  the  same  thing  whether  we  suppose  the  weights  to 
move  over  the  beam,  or  suppose  the  weights  stationary  and  the 
beam  to  move.  In  either  case  we  obtain  every  possible  posi- 
tion of  every  weight  relatively  to  the  ends  of  the  beam. 

The  severest  load  to  which  we  can  subject  a  railway  truss,  for 
example,  is  when  the  span  is  filled  with  locomotives.  If  we 
suppose,  for  illustration,  in  round  numbers,  the  distance  between 
the  three  axles  of  the  locomotive  3  ft.  6  in.,  between  the 
axles  of  the  tender  5  ft.  6  in.,  between  the  foremost  tender 
and  the  back  locomotive  axle  4  ft.,  and  the  entire  length  of 
locomotive  and  tender  34  ft.  6  in.,  and  then  suppose  the  weight 
upon  each  locomotive  axle  13  tons,  and  upon  each  tender  axle 
8  tons,  we  have  a  system  of  weights  in  fixed  order  and  at  fixed 
distances,  and  the  truss  should  be  investigated  for  a  series  of 
these  systems,  as  many  as  can  be  placed  upon  the  span,  passing 
over  it  from  one  end  to  the  other. 

In  PL  9,  Fig.  32  (a),  we  assume  two  such  locomotives  as 
shown  by  P!_IO,  and  construct  the  force  and  equilibrium  poly- 
gons. The  forces  are  symmetrically  arranged  with  respect  to 
a  central  point,  and  the  pole  in  the  force  polygon  is  therefore 
taken  perpendicular  from  the  middle  of  the  force  line. 

Now  the  system  of  forces  being  as  represented,  suppose  the 
span  to  shift.  Thus  suppose  the  span  of  a  given  length  repre- 
sented by  Sj  Si  in  the  Fig.  Then  0  6  is  the  line  closing  the 
polygon  for  this  position  of  the  span,  and  a  parallel  to  0  6  in 
the  force  polygon,  viz.,  C  L  gives  the  reactions  at  the  ends. 
Let  now  the  span  move  from  Si  Sx  to  s5  s5 ;  we  have  a  new  po- 

*  Elemente  der  Gra/phitchen  StatiJc,  Bauschinger. 


54:  MOMENT  OF  KUPTTJKE   OF   PAEALLEL   FOKCES.      [CHAP.  V. 

sition  for  the  line  closing  the  polygon  and  new  reactions.  As 
the  span  continues  to  shift  to  the  right,  the  lines  closing  the 
polygon  revolve,  and  as  their  projections  are  always  constant, 
viz.,  equal  to  the  span,  they  are  all  tangent  to  &  parabola,  which 
they  therefore  envelop. 

48.  Properties  of  tills  Parabola. — This  parabola  has  sev- 
eral important  properties  which  will  aid  us  in  the  investigation 
of  the  case  above  proposed.*     In  PL  9,  Fig.  32  (d),  let  XX  be 
the  line  along  which  the  span  is  shifted ;  a  M  and  a  N  the 
outer  sides  of  the  polygon,  intersecting  at  a,  along  which  the 
closing  lines  slide  as  they  revolve.     For  a  given  position  s  s  of 
the  span,  or  cr  is  the  corresponding  line.     SQ  s0  is  the  position  of 
the  span,  for  which  the  centre,  C0,  lies  in  the  vertical  through  a. 
In  this  position  cr0  cr0  is  tangent  to  the  parabola  at  o>0,  its  middle 
point,  and  upon  this  line  lie  the  centres  of  all  the  other  lines 
(taken  of  course  as  reaching  from  a  N  to  a  M).     Now  the 
point  of  tangency,  /3,  of  any  other  line,  as  cr  cr,  with  the  parabola, 
is  as  far  from  the  centre  of  that  line,  7,  as  the  centre  of  that 
line  is  itself  from  CQ.     We  have  then  only  to  make  c  b  equal  to 
c  CQ,  and  drop  a  perpendicular  through  1}  to  find  /3.     Thus  for 
the  position  ^  st  and  the  line  ^  ar^  to  find  the  point  of  tangency 
c\,  make  c±  dl  equal  to  c±  CQ,  and   draw  d±  c^  perpendicular  to 
intersection  with  cq  ar^ 

Inversely  we  may  find  that  position  for  the  span  s  s,  for  which 
the  vertical  through  a  given  point,  b,  shall  pass  through  the 
point  of  tangency. 

We  have  only  to  move  the  span  so  that  its  middle  point  c 
shall  be  as  far  from  CQ  as  it  is  already  from  the  given  point,  or 
make  c  c0  equal  to  c  b.  (See  Art.  75.) 

If  we  shift  now  the  span  s  s,  and  at  the  same  time  the  point 
ft  through  an  equal  distance,  the  intersections  of  the  vertical 
through  b,  with  the  corresponding  closing  lines  of  the  polygon, 
will  all  lie  upon  the  same  line  a  cr. 

If  therefore  b\  is  such  an  intersection,  b  has  been  moved  from 
b  to  b\,  and  hence  the  span  from  s  s  to  sl  s^. 

49.  Different    Cases    to  be  Investigated. — We  are  now 
ready  to  investigate  the  effect  of  a  live  load  such  as  represented 
in  PI.  9,  Fig.  32  (a).     For  the  determination  of  the  proportions 
of  the  truss  the  following  points  are  specially  important : 

*  See  Memente  der  Graphischen  Statik,  Bauschinger,  pp.  108-114.  Also, 
Die  GrapliiscJie  Statik,  Culmarm,  pp.  136-141. 


CHAP.  V.]      MOMENT    OF    RUPTURE    OF    PARALLEL   FORCES.  55 

1.  When  a  certain  number  of  wheels  pass  over  the  truss,  but 
without  any  passing  off,  or  new  ones  coming  on ;  what  position 
of  the  system  gives  the  maximum  moment  at  any  given  cross- 
section    not  covered    by  the    system,  and    how  great   is  this 
moment? 

2.  Under  the  same  supposition  as  above,  what  position  of  the 
load  gives  the  greatest  moment  for  a  given  point  covered  by  one 
of  the  load  systems  ? 

3.  Among  all  the  various  points  of  the  span,  at  which  is  found 
the  greatest  maximum  moment,  for  what  position  of  the  load 
does  it  occur,  and  how  great  is  it  ? 

4.  If  the  number  of  wheels  is  indeterminate,  how  many  must 
pass  on,  and  what  position  must  they  have  to  give  at  any  point 
the  greatest  maximum  moment;  where  is   the  corresponding 
cross-section,  what  position  must  the  load  have,  and  how  great 
is  this  maximum  moment  ? 

The  three  first  questions  are  easily  solved  by  the  aid  of  the 
above  properties  of  the  parabola,  enveloped  by  the  closing  lines 
of  the  equilibrium  polygon,  corresponding  to  different  positions 
of  the  span. 

Thus,  as  regards  the  first  question,  let  the  given  cross-section 
be  £,  PL  9,  Fig.  32  (df),  and  suppose  the  span  s  s  in  the  position 
where  the  vertical  through  ~b  intersects  a  a  at  the  point  of  tan- 
gency  0.  When  now  the  span  shifts,  the  intersection  of  the 
ordinate  through  5,  with  the  corresponding  tie  line,  will  always 
lie  upon  a  cr.  But  this  ordinate  gives  the  reduced  moments  for 
1)  (reduced  to  pole  distance  H.)  The  greatest  of  these  moments 
will  then  be  simply  the  greatest  of  the  ordinates  between  a  cr 
and  the  polygon,  and  will  always  be  found  at  an  angle  of  the 
same.  When  found,  we  have  at  once  the  position  of  t>,  and  of 
course  of  the  span  with  reference  to  the  given  loads.  This  is 
always  such  that  a  wheel  stands  over  the  given  section. 

Thus  in  Fig.  32  (a\  supposing  the  *  four  wheels  P6  to  P9  to 
pass  over  the  span  t±  ^,  we  seek  the  position  of  the  load  to  give 
the  greatest  moment  at  a  point  £  of  the  span  from  the  left, 
therefore  -g-th  from  the  middle. 

We  lay  off  the  span  in  such  a  position,  ^  tly  that  its  centre  is 
distant  from  the  intersection  a  of  the  outer  lines  of  the  poly- 
gon by  Jth  of  the  span. 

The  ordinate  through  the  given  point  now  passes  through  the 
point  of  tangency  of  the  tie  line  and  parabola.  We  draw  this 


56  MOMENT   OF   KUPTURE   OF   PARALLEL   FORCES.      [CHAP.  Y. 

tie  line  ^  9,  and  seek  the  greatest  ordinate  between  it  and  the 
polygon.  This  we  find  at  7,  and  directly  above  7  the  given 
point  must  lie,  and  hence  we  have  the  position  of  the  span,  viz., 
1 1.  If  the  scale  of  tons  is  ten  tons  to  an  inch,  of  distance  5  ft. 
to  an  inch,  and  the  pole  distance  H  is  assumed  12^  ft.  =  2-J 
inches,  the  scale  of  moments  will  be  10  x  2.5  x  5,  —  125  ft,  tons 
to  an  inch. 

As  to  the  second  question;  the  position  of  the  span  required, 
is  that  where  the  vertical  through  the  given  point  of  the  system 
S  Fig.  32  (a),  intersects  the  corresponding  tie  line  at  its  point 
of  tangency  with  the  parabola ;  all  other  tie  lines  intersect  this 
vertical  in  a  point  between  the  tangent  point  and  the  polygon. 
The  middle  of  the  span  must  then  lie  midway  between  the  in- 
tersection a  of  the  outer  polygon  sides  and  the  point  s,  where 
the  vertical  through  S  meets  the  line  X  X.  Thus  the  span  has 
the  position  t2  t2. 

The  third  question,  finally,  is  easily  solved  if  the  parabola  en- 
veloped by  the  tie  lines  is  drawn.  The  greatest  ordinate  be- 
tween this  parabola  and  the  polygon  gives  the  greatest  moment, 
and  the  point  and  the  position  of  span  required,  since  the 
middle  of  the  span  must  be  half-wray  between  the  point  given 
by  this  ordinate  and  a. 

The  greatest  moment  is  always  found  upon  an  ordinate 
through  an  angle  of  the  polygon. 

If,  however,  the  parabola  is  not  drawn,  we  find  by  trial  at 
several  angles,  drawing  the  tie  lines  and  comparing  the  corre- 
sponding ordinates,  the  ordinate  required.  Here  the  following 
considerations  may  aid : 

When  the  load  is  uniformly  distributed,  the  maximum  mo- 
ment is  in  the  middle  of  the  span,  and  at  the  same  time  in  the 
vertical  through  the  intersection  a  of  the  outer  polygon  sides. 
The  polygon  itself  becomes  a  parabola.  The  less  uniform  the 
load  is.  the  more  this  point  approaches  the  heaviest  loaded  side, 
as  also  the  intersection  a,  though  not  in  the  same  degree.  For 
loads  not  exceedingly  unsymmetrical  the  point  may  be  sought 
for,  then,  in  the  neighborhood  of  a,  i.e.,  near  the  resultant  of  the 
forces  acting  upon  the  truss.  Thus  in  our  example  we  are  jus- 
tified in  selecting  the  corner  7  of  the  polygon,  nearest  the  point 
of  intersection  a. 

5O.  Mo§t  unfavorable  Position  of  Load  upon  a  Beam  of 
given  Spun. — The  fourth  question  above  requires  a  somewhat 


CHAP.  V.]  PARALLEL   FOECES.  57 

more  extended  consideration.  The  most  unfavorable  position 
of  a  system  of  given  concentrated  forces  is  when  it  causes  the 
greatest  moment  at  the  cross-section  of  rupture.  This  position 
is  from  the  preceding,  given  by  taking  the  centre  of  the  beam 
midway  between  the  vertical  through  the  point  of  intersection 
of  the  outer  sides  of  the  equilibrium  polygon  and  the  nearest 
angle  of  the  same.  If  with  this  centre  we  increase  the  span, 
the  maximum  moment  increases  until  the  span  has  the  greatest 
length  possible  without  more  wheels  coming  on. 

Thus  for  the  two  wheels  P4  and  P3,  PL  9,  Fig.  32  (a),  a  is 
the  intersection  of  the  outer  polygon  sides,  and  4  the  nearest 
polygon  angle.  The  almost  equally  near  angle  5  gives  at  any 
rate  no  greater  moment.  In  order  then  that  these  two  weights 
may  cause  the  greatest  maximum  moment,  the  middle  of  the 
beam  must  lie  half-way  between  a^  and  4 ;  and  as  the  span 
increases  in  length  this  moment  increases,  and  is  then  greatest 
when  the  span  reaches  to  $i  or  P3. 

If  now  the  span  still  increases  so  as  to  also  include  P3,  the 
point  of  intersection  of  the  outer  polygon  sides  recedes  to  a2, 
where  in  our  Fig.  it  coincides  almost  exactly  with  the  polygon 
angle  4.  Here  then,  approximately  at  4,  we  must  locate  the 
centre  of  the  beam.  If  we  take  the  same  length  of  span  a£ 
before,  that  is,  make  the  half  span  0%  s2  equal  to  the  distance 
from  Si  to  the  point  midway  between  a^  and  4,  we  see  by  draw- 
ing the  closing  lines  for  these  two  positions  of  the  span,  that 
the  maximum  moments  measured  upon  the  vertical  through  4 
are  almost  exactly  equal  in  each  case.  For  a  smaller  length  of 
span  including  the  three  weights,  the  maximum  moment  de- 
creases, and  is  less  therefore  than  the  maximum  moment  already 
caused  by  the  t\vo  wheels.  The  span  ^  ^  may  then  be  regarded 
as  the  greatest  for  which  the  two  wheels  P4  P5  give  the  greatest 
possible  maximum  moment.  As  the  space  s2  &>,  upon  which  we 
have  now  three  wheels,  increases,  the  moment  increases,  and  is 
greatest  when  the  span,  its  centre  always  remaining  now  at  «2 
reaches  to  s'2  or  to  P2. 

If  now  it  still  increases  so  as  to  also  include  P2,  the  intersec- 
tion of  the  outer  polygon  sides  retreats  to  a%.  The  nearest 
polygon  angle  is  still  4,  and  midway  then  between  a$  and  4  we 
must  now  locate  the  middle  of  the  beam.  If  from  this  centre 
we  lay  off  the  half  span  equal  to  a%  s'%,  to  s3,  and  draw  the  clos- 
ing line  for  this  position  of  the  span,  we  see  as  before  that  the 


58  MOMENT   OF   KUPTFRE.  [CHAP.  V. 

moment  given  by  the  ordinate  at  4  is  for  either  case  almost 
exactly  the  same.  Any  less  span  including  the  four  weights 
would  give  a  less  moment ;  less,  therefore,  than  the  moment 
already  caused  by  the  three  weights.  The  span  s'2  s'2  then 
precisely  as  before,  is  the  extreme  limit  upon  which  the 
three  wheels  P3  to  P5  cause  the  greatest  possible  maximum 
moment. 

In  a  precisely  similar  manner  we  find  that  the  span  s'3  s's 
with  a  centre  midway  between  a3  and  4  is  the  limiting  span  for 
the  four  wheels  P2  to  P5. 

If  now  the  span  still  increases  so  that  P!  comes  on,  the  inter- 
section of  the  outer  polygon  sides  falls  in  our  Fig.  nearly  at  s1? 
and  since  this  point  also  happens  to  correspond  almost  exactly 
with  the  angle  3,  we  take  the  centre  of  the  beam  at  s^.  The 
greater  the  span  now  becomes,  the  greater  the  maximum 
moment.  The  greatest  length,  however,  which  the  span  can 
have  without  including  P6,  is  twice  ^  6,  or  twice  the  distance 
between  ^  and  P6.  If  P0  also  comes  on,  the  intersection  of  the 
polygon  sides  is  found  at  «5,  and  the  nearest  polygon  angle  is  4. 
Midway  then  between  a5  and  4  is  the  new  centre  of  the  beam, 
while  before  P6  came  on,  it  was  nearly  at  s±.  But  for  centre 
st  the  half  span  was  Si  6,  while  now  it  is  somewhat  less  than 
4  6  ;  therefore  considerably  smaller.  Since,  however,  we  wish 
to  follow  the  span  as  it  continues  increasing,  we  must  compare 
those  two  spans  which  are  equal  before  and  after  the  coming 
on  of  P6.  The  right-hand  ends  of  these  spans,  viz.,  s'4  and  s5 
must  evidently  be  distant  each  side  of  6,  by  the  half  distance 
of  their  centres  sl  and  4,  or  0%  (more  accurately  the  point  half- 
way between  a5  and  4,  but  a5  and  4  lie  in  our  Fig.  so  nearly  to- 
gether that  the  centre  cannot  be  indicated  more  exactly).  We 
make  then  ^  «'4  =  o^  =  Mx  6,  provided  that  ML  is  taken  half- 
way between  the  centre  ^  and  0%. 

An  exact  construction  shows  that  the  maximum  moments  for 
these  two  spans,  the  one  given  by  the  ordinate  through  3,  the 
other  by  the  ordinate  through  4,  are  almost  exactly  equal,  and 
moreover,  that  the  maximum  moment  for  the  span  Si  Sl  of  equal 
length  whose  centre  is  at  Mt  is  also  almost  exactly  equal,  when 
measured  upon  the  vertical  through  M!.  We  can  therefore 
take  Si  St  as  the  limit  of  those  spans  for  which  the  five  wheels 
Pt  to  P5  cause  the  greatest  maximum  moment. 

Taking  on  now  the  seventh  wheel,  the  intersection  of  the 


CHAP.  V.]  PARALLEL   FORCES.  59 

outer  polygon  sides  is  at  a6  and  the  nearest  polygon  angle  is  5. 
Half-way  between  a6  and  5  we  must  then  take  the  centre,  while 
before  it  lay  at  a^  (nearly).  If  we  take  then  M2  half-way  be- 
tween 0-2  and  this  new  centre,  we  find  precisely  as  before  the 
span  S2  S2  with  centre  M2,  and  right  end  at  P7,  as  the  limiting 
span  for  the  six  wheels  Pl  to  P6.  The  same  holds  good  for  the 
span  S3  S3  with  centre  M3,  for  the  seven  wheels  Pi  to  P7,  and 
so  on.  If,  according  to  supposition,  P8  P4  P5  are  3  ft.  6  in. 
apart,  P2  and  P3  4  ft.,  and  P!  and  P2  5  ft.  6  in.  apart;  then  for 
spans  up  to  ^  SL  =  say  8  ft.,  the  two  wheels  P4  P5  will  give  the 
greatest  maximum  moment,  and  their  place  upon  the  beam  is 
given  by  the  position  of  the  centre  (half-way  between  a^  and  4). 
From  about  8  ft.  to  15  ft.  span,  or  s2  s2  the  three  wheels  P3  to  P5 
give  the  greatest  maximum  moment,  and  the  centre  of  the  span 
is  located  at  a%.  For  spans  from  1 5  ft.  to  19  ft.  span,  or  s'z  8\, 
the  four  wheels  P2  to  P5  give  the  maximum  moment,  and  the 
centre  is  at  ^ ;  and  so  on.  Thus  for  a  span  of  any  given  length 
we  have  at  once  the  weights  and  their  position,  in  order  to 
cause  the  greatest  maximum  moment,  as  also  the  -place  of  this 
moment,  viz.,  the  point  vertically  over  that  angle  of  the  equi- 
librium polygon  nearest  the  centre  of  the  span.  The  ordinate 
through  this  point  included  by  the  equilibrium  polygon,  and 
the  closing  line  for  the  given  span,  taken  to  the  moment  scale 
gives  this  moment  at  once  ;  or  this  ordinate  taken  to  the  scale 
of  force  must  be  multiplied  by  the  previously  assumed  jpale 
distance. 

51.  Oreate§t  Moment  of  Rupture  caused  by  a  Sy§tem  of 
Moving  ILoads  at  a  given  Cross-Section  of  a  Beam  of  given 
Span. — For  beams  or  trusses  of  long  span,  which  are  as  a  rule 
caused  to  vary  in  cross-section,  it  is  not  sufficient  merely  to  find 
the  greatest  maximum  moment  which  a  given  system  of  con- 
centrated forces  can  cause  ;  we  must  also  know  for  a  number 
of  individual  cross-sections,  the  maximum  moments  which  can 
ever  occur. 

For  this  purp6se  the  force  and  equilibrium  polygons  being 
first  constructed,  we  shift  as  above  the  given  span  along  a 
horizontal  line,  and  draw  for  each  successive  position  of  the 
span  the  corresponding  closing  line  in  the  equilibrium  polygon, 
marking  the  point  where  each  closing  line  is  intersected  by  a 
vertical  through  the  given  cross-section,  which  of  course  moves 
with  the  span,  keeping  always  the  same  position  with  reference 


60  MOMENT   OF   RUPTURE.  [dlAP.  V. 

to  the  ends.  The  points  thus  obtained  form  a  curve,  and  the 
greatest  ordinate  between  this  curve  and  the  polygon  gives  the 
greatest  moment  which  can  act  at  the  given  cross-section. 
This  greatest  ordinate  will  always  be  found  at  an  angle  of  the 
polygon,  and  hence  a  weight  must  always  rest  upon  the  cross- 
section.  Since  the  cross-section  itself  must  lie  upon  this  ordi- 
riate,  we  have  directly  the  position  of  the  span  with  reference 
to  the  given  forces.  The  closing  line  for  this  position  being 
then  drawn,  a  parallel  to  it  in  the  force  polygon  gives  the  reac- 
tions for  this  position. 

The  reader  will  do  well  to  make  the  construction  indicated 
for  an  assumed  span  and  system  of  weights,  to  convenient  scales, 
checking  the  results  by  computation.* 

The  above  method  applies  more  particularly  to  solid  or 
"plate  "  girders,  beams,  or  trusses.  It  may  of  course  be  applied 
to  framed  structures  also,  such  as  those  illustrated  in  chapter 
first.  Thus  the  moment  at  any  point,  divided  by  the  depth  of 
truss  at  that  point,  gives  the  strain  in  flanges.  The  more  pre- 
ferable, as  perhaps  also  the  simplest  method  of  determining  the 
strains  in  such  cases,  however,  is  to  find  the  reactions  due  to 
each  individual  weight.  Each  reaction  can  then  be  followed 
through  the  structure,  as  explained  in  that  chapter,  and  the 
strains  in  every  member  for  every  weight  in  every  position  can 
thus  be  obtained  and  tabulated.  An  inspection  of  the  table 
\^11  then  give  at  once  the  strains  due  to  the  united  action  of 
any  desired  number  of  these  weights. 

We  have  thus  two  methods  for  the  solution  of  such  cases ; 
first,  by  the  composition  and  resolution  of  forces,  and,  second, 
by  the  equilibrium  polygon  and  moments  of'  rupture,  and  may, 
if  we  choose,  check  the  results  obtained  by  one  method  by  the 
other.  In  most  practical  cases  involving  framed  structures, 
however,  the  first  method  is  preferable  as  being  simpler,  quicker 
of  application,  and  of  superior  accuracy. 

For  solid-built  beams  or  "plate  girders,"  etc.,  the  second 
method  comes  more  especially  into  play.  The  determination 
of  the  strains  in  a  structure  of  this  kind  from  the  known  mo- 
ment of  rupture  at  any  point,  requires  a  knowledge  of  the 
moment  of  inertia  of  the  cross-section  at  that  point,  and  this 
may  also  be  found  by  the  Graphical  method. 

*  This  construction  is  given  in  Art.  15,  Fig.  VIII.,  of  the  Appendix. 


CHAP.  VI.~|  MOMENT   OF   INERTIA.  61 


CHAPTER  VI. 


MOMENT    OF   INERTIA   OF   PARALLEL    FORCES. 

52.  THUS  far  we  have  seen  that  by  the  graphic  method  we 
can  in  any  practical  case  determine  the  moment  of  the  exterior 
forces  acting  upon  a  piece  at  any  cross-section  of  that  piece. 
But  the  exterior  forces  give  rise  to  and  are  resisted  by  molecu- 
lar or  interior  forces.  Now  the  moment  of  the  exterior  forces 
being  found,  the  cross-section  of  the  piece  at  any  point  being 
known,  and  one  of  the  dimensions  of  this  cross-section  being 
assumed,  it  is  required  to  find  the  other  dimension,  so  that  the 
strain  per  unit  of  area  of  cross-section  shall  be  less  than  the 
recognized  safe  strain  of  the  material  as  found  by  experiment. 

The  moment  of  the  exterior  forces  at  any  cross-section  we 
call  the  moment  of  rupture  ;  and  designate  it  by  M.  Let  d  = 
the  depth  of  cross-section.* 

y  =  the  variable  distance  of  any  fibre  above  or  below  the 
neutral  axis. 

/3  =;  'the  breadth  of  the  section  at  the  distance  y  from  the 
neutral  axis,  and  consequently  a  variable,  except  in  the  case  of 
rectangular  sections. 

s  =  the  horizontal  unit  strain  exerted  by  fibres  in  the  cross- 
section  at  a  given  distance  c  from  the  neutral  axis. 

o 

Then  since  the  fibres  exert  forces  which  are  proportional  to 
their  distance  from  the  neutral  axis  or  to  their  change  of  length, 
the  unit  strain  in  any  fibre  at  a  distance  y  from  the  neutral 

o   n  i 

axis  will  be  — — .     Let  the  depth  of  this  fibre  be  d  y,  then,  since 
c 

the  breadth  of  section  is  /9,  the  total  horizontal  force  exerted 

o 

by  the  fibres  in  the  breadth  /?,  will  be  —  f3y  dy.    The  moment 

C 

Q 

of  this  force  about  the  neutral  axis  will  be—  /3  y*  d  y,  and  the 

c 


Thewy  of  Strains,  Stoney,  p.  43,  Art.  07. 


62  MOMENT   OF   INERTIA.  [CHAP.  VI. 

integral  of  this  quantity  will  be  the  sum  of  the  moments  of  all 
the  horizontal  elastic  forces  in  the  cross-section  round  the  neu- 
tral axis,  that  is,  equal  to  the  moment  of  rupture  of  the  section 
in  question.  We  have  therefore 


For  a  rectangular  cross-section,  for  instance,  (B  is  constant 
and  equal  to  the  breadth  b.     Representing  the  depth  by  d  we 

have  M  =  y^  —  ,  or  if  we  make  c  the  distance  of  the  extreme 

d 

fibres  —  - 


from  which  M  being  known,  as  also  s,  if  we  assume  b  we  can 
find  d  or  the  reverse. 

The  integral  //?  y*  dy  is  the  moment  of  inertia  of  the  cross- 

section,  and  may  be  defined  as  the  sum  of  the  products  obtained 
~by  multiplying  the  mass  of  each  elementary  particle  by  the 
square  of  its  distance  from  the  axis.  [See  Supplement  to  Chap- 
terVIL.Art.  10.] 

From  the  above,  we  see  its  importance  in  determining  the 
strain  at  any  distance  from  the  neutral  axis,  or  in  proportioning 
the  cross-section,  so  that  the  resulting  strain  shall  be  less  than 
a  given  quantity  at  any  point.  We  see  also  that  for  a  rectan- 

~b  dz 
gular  cross  -section  the  moment  of  inertia  is  -^-,  where  b  is  the 

breadth  and  d  the  depth. 

53.  Graphical  Determination.  —  "We  have  already  seen 
that  the  moment  of  a  force,  as  Pt  (PI.  6,  Fig.  20)  with  reference 
to  any  point,  as  0,  is  given  by  the  ordinate  n  m  multiplied  by 
the  constant  H  (Art.  38).  The  ordinate  n  m  then  represents 
the  product  of  Pt  multiplied  by  the  horizontal  distance  of  b 

from  n.     But  the  area  of  the  trianle  ~b  n  m\$>mnK      b  n  = 


PI  x  -^  b  n*,  that  is,  the  area  of  the  triangle  b  n  m  represents 

one-half  the  moment  of  inertia  ofl?i  with  respect  to  o.  Just 
as  the  exterior  ordinates  of  the  equilibrium  polygon  have  been 
shown  to  have  a  certain  significance,  and  to  represent  the  mo- 


CHAP.  VI.]  MOMENT   OF   INEETIA.  63 

ments  of  the  forces,  so  the  exterior  areas  of  the  equilibrium 
polygon  represent  the  moments  of  the  moments,  or  the  moments 
of  inertia.  Thus  in  PI.  8,  Fig.  30,  the  exterior  parabolic  area 
oC/i  should  be  one-half  the  moment  of  inertia  of  the  rectangle 
or  load  area  o  p  r  h. 

Let  us  see  if  this  is  so.     The  area  of  the  triangle  o  h  C  is  — 

o  h  x  the  ordinate  S  C.  This  ordinate  S  C  gives,  as  we  have 
seen,  the  moment,  with  respect  to  S,  of  the  reaction.  "We  can 
therefore  find  its  value.  Thus  if  p  is  the  load  per  unit  of  length, 

and  I  is  the  length,  ^~  is  the  reaction,  and  ~—  this   moment. 

I      pi2       p  1B 
The  area  of  the  triangle  o  c  h  is  therefore  -~  X  —r—  =  --— . 

The  parabolic  area  odh  is  f  of  the  circumscribing  rectan- 
gle. This  rectangle  is  I  x  S  d.  The  ordinate  S  d  is  equal  to 
S  C  —  dC.  We  have  already  found  SC  and  dC  is  the  sum  of  the 

7}  I       I*         'D  1?  T)  ft 

moments  of  P!  and  P2,  or  ~*  x  -.  —  ^r~.     Hence  S  d  —  -—-  — 

A  4:  O  4: 

p  p        p  p 

^—  =    Q~~~-     The  area  °f  *ne  circumscribing  rectangle  is  then 

7}  1?  2  V  1?  T)  1? 

£p  Two-thirds  of  this  is  -~9  which  subtracted  from  &&- 
gives  for  half  the  moment  of  inertia  ~-j  p  Z8.  Hence  the 

2i^c. 

moment  of  inertia  is  —  p  P,  as  should  be. 

54.  We  see  therefore  the  significance  of  the  area  of  the  equi- 
librium polygon. 

If,  when  a  number  of  forces  are  given,  we  form  the  force 
polygon,  and  then  the  equilibrium  polygon,  the  ordinates  to 
this  last  give  the  moments  to  the  assumed  pole  distance.  If 
now  we  take  these  moments  themselves  as  forces  applied  at  the 
same  points,  form  a  new  force  polygon  with  new  pole  distance, 
and  new  equilibrium  polygon,  the  ordinates  to  this  new  polygon 
to  the  new  pole  distance  will  give  the  moments  of  the  moments 
or  the  moments  of  inertia  of  the  forces.  The  same  method  is 
applicable  to  moments  of  a  higher  order,  but  in  practice  we 
have  only  to  do  with  those  of  the  second  order  alone. 

55.  Radius  of  Gyration. — The  moment   of   inertia  of   a 
system  of  parallel  forces  P!  P2  etc.,  in  a  plane,  with  reference 


64  MOMENT    OF   INERTIA.  [dlAP.  VI. 

to  an  axis  from  which  the  points  of  application  are  distant  ql  $>, 
etc.,  is  then  2  P  <f.  This  is  the  product  of  three  .quantities, 
one  of  which  is  measured  by  the  scale  of  force,  and  the  other 
two  by  the  scale  of  length.  We  can  therefore  regard  it  as  the 
product  of  the  square  of  a  certain  length  by  the  sum  of  the 
given  forces,  or  2  P  <f  =  P  2  P.  We  call  k  the  radius  of 
gyration. 

In  order  to  find  the  moment  of  inertia  of  a  system  of  parallel 
forces  then,  we  must  by  the  preceding  Art.  construct  two  force 
and  equilibrium  polygons.  If  the  pole  distances  are  H  and  H', 
and  the  segments  into  which  the  axis  is  divided  by  the  produced 
sides  of  the  polygons  are  P\  P'2  and  P'^  P"2  etc.,  respectively, 
then 

2  P  (f  =  H  H'  2  P" 
and  the  radius  of  gyration  is  given  by 

HH'  2P"  =  L*  2P 

, 
or  Js-. 


This  expression  is  easy  to  construct.  Thus  for  example  in 
PI.  11,  Fig  33,  let  o  n  C  be  the  first  force  polygon,  o  n  the  force 
line,  containing  the  forces  P  ;  C  the  pole,  and  H  the  pole  dis- 
tance. Make  o  b  equal  to  the  second  pole  distance  H/  and  draw 
b  c  parallel  to  n  c  and  c  t  parallel  to  H.  Then 

HH' 


whence  k  = 

If,  therefore,  in  Fig.  33  (&),  m"0  m"n  is  the  segment  of  the 
axis  cut  off  by  the  outer  sides  of  the  second  equilibrium  poly- 
gon, that  is,  if  m"0  m"n  —  -2*  P",  we  have  only  to  prolong  m"0 
m"n  to  L,  making  m"  L  =  A,  and  describe  a  semicircle  upon 
m"0  L,  and  erect  the  perpendicular  m"  7t,  which  will  be  equal 
to  %.  In  general,  the  pole  distance  H  and  H'  can  be  taken  ar- 
bitrarily, but  it  is  often  advantageous  to  take  H  (sometimes  H' 
also)  equal  to  2  P.  Then 


We  should  then  have  in  Fig.  33  simply  to  increase  m"''0  m"^ 
by  the  second  pole  distance  H',  and  then  proceed  as  above  to 
find  L 


CHAP.  VI.]  MOMENT   OF   INEKTIA.  65 

It  is  to  be  remembered  that  q±  q^  etc.,  the  distances  of  the 
points  of  application  of  the  forces  from  the  axis,  may  be  meas- 
ured in  any  direction,  and  H  is  parallel  to  this  direction,  and 
is  not  therefore  necessarily  perpendicular  to  o  n. 

The  above  will  be  rendered  plain  by  reference  to  Fig.  34-, 
PI.  10.  We  suppose  four  forces  applied  at  the  points  At  A2  A3 
A4  respectively,  and  acting  parallel  to  XX.  Required  the  mo- 
ment of  inertia  of  these  forces  and  the  radius  of  gyration,  the 
distances  gl  q^  etc.,  being  measured  parallel  to  Y  Y.  First  we 
form  the  force  polygon  by  laying  off  along  X  X,  0  1,  1  2,  2  3, 
3  4,  parallel,  and  in  the  direction  of  action  of  the  forces,  cligos- 
ing  a  pole  C,  and  drawing  C  0,  C  1,  C  2,  etc.  We  now  construct 
the  corresponding  equilibrium  polygon,  CI,  III,  II  III,  in  IV, 
etc.  The  segments  01',  1'2',  2'  3',  etc.,  represent  the  statical 
moments  of  the  forces  with  reference  to  X  X.  That  is,  these 
segments  to  the  scale  of  force  multiplied  by  the  pole  distance  C  y 
parallel  to  Y  Y  to  the  scale  of  distance,  give  the  statical  moments 
of  the  forces.  Now  we  take  these  segments  themselves  as  forces, 
and  suppose  them  acting  at  the  former  points  of  application. 
With  the  same  pole  as  before  we  draw  CO,  C  1',  C  2',  etc.,  and 
form  the  corresponding  equilibrium  polygon  CI,  III',  II  III'',  etc. 
The  sum  of  the  segments  of  XX  cut  off  by  the  outer  lines  of  this 
polygon,  or  o  y,  to  the  scale  of  force  multiplied  by  H  H'  or  C  if 
gives  the  moment  of  inertia  of  the  forces  with  respect  to  XX. 
This  moment  then  is  M  =  0  y  x  Cy* 
where  Oy  "=  2P"  and"Cp  =  HH'. 

The  radius  of  gyration  Jc  is,  as  we  have  seen,  given  by 


CT  rrr  ^  T)fr 

,  5P  being  equal  to  0  4  in  the  Fig.    Hence 


04 

If,  then,  we  lay  off  0  d  =  0  4,  and  make  0  c  =  C  y,  and  make 
the  angle  dee  a  right  angle,  we  shall  find  a  point  e  to  the  right 

/"I    ~  2  TT  TT/ 

and  Oe  will  be  equal  to  --  =  -y-     Upon  ey  now  describe 


a  semi-circle,  the  point  of  intersection  y  with  the  perpendicular 
through  0  will  give  (Art.  55) 

/Oy  x  Cy3          /HH'SP" 
=  V  "     Q4   c      =  V  "  ^£P  —  =  ^  —  radius  of  gyration. 

The  square  of  this  line,  then,  multiplied  by  2  P  or  0  4,  will  give 
5 


66  MOMENT   OF   INERTIA.  [CILAP.  VI. 

at  once  the  moment  of  inertia  of  the  given  four  forces  with 
reference  to  X  X  and  Y  Y  as  axes.  If  we  were  to  suppose  the 
same  forces  with  the  same  points  of  application  to  act  parallel 
to  Y  Y  instead  of  X  X,  the  distances  qt  q2  being  measured  par- 
allel to  X  X  instead  of  Y  Y,  we  should  have  the  force  polygon 
G!  O  li  %i  3t  4X  instead  of  001234,  and  a  precisely  similar 
construction  would  give  us  o  x  multiplied  by  pole  distance  for 
the  moment  of  inertia,  and  0  a'  for  the  radius  of  gyration.  We 
recommend  the  reader  to  follow  through  the  construction  as 
shown  by  Fig.  34. 

56.  Curve  of  Inertia—  Ellipse  and  Hyperbola  of  Inertia. 
—  II  having  found  the  radius  of  gyration  as  above,  we  lay  it  off 
from  the  axis  on  either  side,  in  a  direction  parallel  to  the  direc- 
tions in  which  <2i  $2?  etc.,  are  supposed  measured,  and  through 
the  points  thus  determined  draw  two  parallels  to  the  axis  M' 
and  M"  on  either  side,  and  then  suppose  the  axis  to  revolve  in 
the  plane  of  the  forces  about  any  point  as  O  situated  in  the 
axis  ;  the  lines  M'  and  M"  also  revolve  and  enclose  a  curve  of 
the  second  degree,  whose  centre  coincides  with  O.  Thus,  if  in 
PL  10,  Fig.  34,  we  lay  off  O  I  along  Y  Y  on  both  sides  of  X  X 
equal  to  o  V  =  k  already  found,  and  then  let  X  X  revolve 
about  O,  K  J  and  J  K  will  also  revolve,  and  enclose  either  an 
ellipse  or  hyperbola. 

In  order  to  prove  this,  take  O  as  an  origin  of  co-ordinates. 
Let  the  co-ordinates  of  the  points  of  application  of  the  forces  Ax 
A2,  etc.,  be  xt  y},  a?2  y»  etc.  From  each  of  these  points  A  draw 
parallels  to  the  axis  of  y,  intersecting  the  axis  of  x  in  the  points 
C.  Then  O  C  =  a?,  A  C  =  y.  Now  pass  through  the  point  O 
an  axis  of  moments  M  in  any  direction,  and  project  for  each 
point  OCA  parallel  to  this  axis  upon  the  line  q,  which  meas- 
ures the  distance  of  each  point  from  the  axis  of  moment  (not 
necessarily  perpendicular  distance).  This  projection  is  evi- 
dently equal  to  q.  Denote  by  a  and  /3  the  ratios  by  which  dis- 
tances along  X  and  Y  must  be  multiplied,  in  order  to  obtain 
their  projections  upon  £,  by  lines  parallel  to  M.  Then 

q=a%+  @y 
for  each  point  of  application,  and  hence 


or  since  for  one  and  the  same  axis  M,  and  direction  £,  a  and 
are  constant, 


CHAP.  VI.]  MOMENT    OF   INERTIA.  67 

In  this  expression  a  and  /9  will  vary  with  the  position  of  M 
and  the  direction  of  ^,  but  2  P  a?2,  2  P  y3  remain  unchanged. 
These  last  expressions  are,  however,  nothing  more  than  the  mo- 
ments of  the  second  order  (moments  of  inertia)  of  the  given 
force  system  with  reference  to  the  co-ordinate  axis,  the  distances 
of  the  points  of  application  being  measured  in  the  direction  of 
the  axis.  They  are  known  if  the  force  system  is  given  and  the 
co-ordinate  system  assumed. 

If  we  put  2  P  ar1  =  a2  2  P,  2  P  f  =  tf  2  P,  2  P  x  y  =  f2  2  P, 
£  and  #,  etc.,  are  the  radii  of  gyration  of  the  moments  of  inertia 
with  reference  to  x  and  y,  and  the  above  eqiiation  becomes 
^P  (f  =  2  P  [a2  a?  +  /32  &2  +  2  a  /9/2] 

If  we  conceive  for  the  assumed  position  of  M,  the  radius  of 
gyration  "k  to  be  found,  and  M'  and  M"  drawn  on  either  side 
at  a  distance  ±  &,  measured  parallel  to  £,  and  indicate  the  dis- 
tances cut  oft'  by  these  lines  from  the  co-ordinate  axes  by  ±  xe 
±  2/e,  and  then  project  these  distances  parallel  to  M  upon  the 
direction  of  q>  or  &,  we  have  Jc  =  a  xe  =  ft  y^  whence 

K          Q          K 

a  —  —      p  =  — 

®e          y* 

and  these  values  substituted  in  the  above  equation  give 


where  ^  is  essentially  positive  in  the  second  term. 
Hence, 


If  we  suppose  the  axis  M  to  change  its  position  revolving  about 
O,  the  segments  xe  y&  cut  off  from  the  axes  of  x  and  y  by 
M7  and  M"  alone  will  change  in  this  equation.  It  is  therefore 
the  equation  of  the  curve  enclosed  by  M7  M".  If  this  curve  is 
known  for  a  given  force  system,  then  the  moment  of  inertia  for 
any  axis  passing  through  its  centre  is  easily  found.  We  have 
only  to  draw  parallel  to  the  axis  two  tangents  to  this  curve,  one 
on  either  side,  and  measure  their  distance  from  M,  in  the  direc- 
tion in  which  the  distances  q  of  the  points  of  application  from 
the  axis  are  taken.  This  distance  is  the  radius  of  gyration, 
and  the  moment  of  inertia  is  simply  the  product  of  its  square 
by  the  algebraic  sum  of  the  forces. 


68  MOMENT   OF   INERTIA.  [CHAP.  VI. 

We  call  the  curve  represented  by  the  above  equation  there- 
fore, the  curve  of  inertia.  If  we  refer  the  curve  to  co-ordinate, 
axes  which  coincide  with  the  conjugate  diameters,  the  equation 
becomes 

^+I2  =  ±i 

o      '  o  -    —1—    *• 

a?      y2 

where  x  and  y  are  the  new  ordinates,  and  A,  B,  the  conjugate 
semi-axes  of  the  curve.  A  and  B  are  therefore  the  radii  of 
gyration  of  the  force  system,  measured  in  the  direction  of  the 
co-ordinate  axes,  and  hence 


where  x  and  y  are  the  co-ordinates  of  the  points  of  application 
of  the  given  forces. 

.  Since  2  P  <f  =  Z?  2  P  if  the  sign  of  -£P  <f  is  the  same  as 
2  P,  J£  is  positive.  When,  on  the  other  hand,  these  signs  are 
different,  #*  is  negative.  That  is,  when  all  the  forces  act  in 
the  same  direction  W  is  positive,  and  we  have 

A2        "R2 
—  +        —  1 

X*  +  y*- 

which  is  the  equation  of  an  ellipse. 

If,  however,  the  parallel  forces  act  in  different  directions,  kt 
may  be  positive  or  negative.  For  cases  where  ^  is  negative, 
either  A2  or  B2  will  be  negative,  and  we  shall  have 

' 


or 


Both  cases  coincide.  The  double  curve  consists  of  two  hyper- 
bolas with  common  assymptotes,  common  centre,  and  equal 
semi-axes.  For  every  axis  M  passing  through  the  common  centre 
O,  we  have  a  pair  of  parallel  tangents  either  to  one  or  the 
other  hyperbola.  The  corresponding  ^  is  positive  for  the  one, 
negative  for  the  other. 

If,  then,  in  the  method  of  construction  to  which  we  shall 
presently  refer,  the  square  of  the  semi-axis  B,  which  lies  in  the 
axis  of  Y,  is  negative,  that  hyperbola  whose  imaginary  axis  lies 
in  Y  gives  ^  positive,  the  other  gives  ^  negative,  and  reversely 
for  the  other  case.  If  the  axis  of  moments  M  coincides  with 


CHAP.  VI.]  MOMENT   OF   INERTIA.  69 

one  of  the  common  assymptotes,  the  radius  of  gyration  and 
moment  of  inertia  with  respect  to  it  of  the  given  force  system 
is  zero. 

57.  Construction  of  the  Curve  of  Inertia. — The  curve  of 
inertia  for  a  given  system  of  parallel  forces  and  given  centre 
O,  is  determined  by  the  direction  of  any  two  conjugate  diame- 
ters, since  as  we  have  seen  in  Art.  55,  PL  10,  Fig.  34,  these 
directions  being  assumed  we  can  find  the  radii  of  gyration  with 
respect  to  XX  and  Y  Y,  and  can  thus  determine  O  a  and  O  5, 
the  semi-diameters.     We  have  then  to  develop  a  principle  by 
means  of  which  these  directions  may  be  determined. 

If  we  denote  the  distances  of  the  points  of  application  of  the 
forces  from  the  axis  of  M  measured  in  any  direction  by  y,  then 
the  statical  moments  of  the  forces,  P  y,  are  indeed  dependent 
upon  the  direction  in  which  y  is  measured,  but  their  relative 
values  remain  the  same.  If  then  being  found  for  any  direction 
of  y,  these  statical  forces  are  considered  as  being  themselves 
parallel  forces  acting  at  the  points  of  application,  and  their 
centre  of  action  is  found  (for  gravity — centre  of  gravity)  for 
some  other  value  of  y,  this  centre  of  action  remains  unchanged. 
For  any  axis  passing  through  this  centre  of  action  the  sum  of 
the  moments  of  the  forces  is  zero.  If  therefore  we  take  a  point 
O  in  the  axis  M  as  origin  of  a  system  of  co-ordinates,  whose 
axis  O  X  may  lie  at  will  in  the  plane  of  the  forces,  while  O  Y 
passes  through  the  centre  of  action ;  the  sum  of  the  moments 
of  the  statical  moments  P  y,  considered  as  forces  acting  at  the 
points  of  application,  with  reference  to  O  Y,  will  be  zero. 
These  moments  however,  provided  that  the  distances  of  the 
points  of  application  are  measured  along  the  co-ordinate  axes, 
are  the  moments  of  'inertia,  viz.,  2  P  y  x.  If  these  are  zero  we 
see  that  the  general  equation  of  the  curve  of  inertia  (1)  Art.  56, 
becomes  that  of  a  hyperbola  referred  to  its  conjugate  diameters 
as  axes.  With  the  centre  O  therefore,  the  line  joining  O  with 
the  centre  of  action,  gives  the  direction  of  the  conjugate  di- 
ameter of  the  curve. 

This  is  the  principle  required.  By  means  of  it  we  can  find 
the  conjugate  diameters  of  the  inertia  curve,  for  a  given  centre 
O,  and  thus  construct  it. 

58.  Con§truetion  of  the  Curve  of  Inertia  for  four  paral- 
lel force§  in  a  Plane.    Example. — As   an   example    let   us 
take    the    four   parallel  forces,  in   PL  10,  Fig.  34,  supposed 


70  MOMENT    OF   INERTIA.  [CHAP.  VI. 

to  act  in  different  directions,  parallel  to  X  X  at  the  points 
A!  Ag,  etc. 

As  before  we  have  the  force  polygon  C  0  1  2  3  4  for  an  arbi- 
trary axis  as  X  X,  and  from  the  corresponding  equilibrium 
polygon,  we  determine  the  statical  moments  with  reference  to 
X  X,  01'  1/2',  etc.,  to  the  basis  C  0.  These  moments  we  again 
consider  as  parallel  forces  acting  at  At  A2,  etc.,  for  which  we 
have  CO  V  2'  3'  4'  and  corresponding  equilibrium  polygon 
C  I  IT  ffl',  etc.  We  then  determine  the  centre  of  action  S,  by 
a  second  polygon  0"  I II"  III",  etc.,  the  sides  of  which  are 
respectively  perpendicular  to  the  first,  according  to  the  process 
for  finding  the  centre  of  gravity,  Art.  30.  The  line  joining  O 
with  S  gives  th&  direction  ofYY,  the  diameter  of  the  curve 
conjugate  to  X  X.  To  find  the  length  of  the  semi-diameters 
0 1}  and  O  #,  we  must  find  the  moments  of  inertia  of  the  forces 
with  reference  to  X  X  and  Y  Y,  taking  the  distances  of  the 
points  of  application  as  measured  parallel  to  these  lines. 

Therefore  instead  of  C  0,  we  must  take  C  y  as  basis  or  pole 
distance,  and  then  find  the  radii  of  gyration  as  already  indi- 
cated in  Art.  55,  viz.,  O  bf  and  O  a' '.  These  distances  laid  off 
along  Y  Y  and  X  X  give  the  semi-conjugate  diameters  of  the 
curve  of  inertia. 

From  the  Fig.  we  see  that  the  force  Pl  whose  direction  from 
left  to  right  we  shall  always  consider  positive,  and  5*  P  —  0  4 
have  the  same  sign.  On  the  other  hand  the  total  moment  of 
inertia  0  y  and  the  moment  of  inertia  of  Pl5  viz.,  0  l"_have  dif- 
ferent signs.  The  square  of  radius  of  gyration  #*  =  ~  y  p — 

is  therefore  negative,  the  radius  itself  or  the  semi-diameter  O  b 
is  imaginary. 

In  similar  manner,  we  see  that  O  a  the  radius  of  gyration  for 
Y  Y  is  real,  since  the  total  moment  of  inertia  O  x  and  S  P  —  0  41? 
have  the  same  signs.  The  curve  is  then  a  double  hyperbola 
with  the  conjugate  semi-diameters  O  a  and  O  b. 

It  is  then  easy  to  find  the  assymptotes  K  K  and  J  J,  and  by 
bisecting  the  angle  which  they  make,  the  principal  axes  A  A 
and  B  B.  In  order  to  find  the  length  of  these  axes,  we  have 
the  well-known  principle  that  for  any  point  as  &,  the  product 
of  a  k  and  k  O  (a  ~k  being  parallel  to  the  assymptote  J  J)  is  equal 

to  -r  the  sum  of  the  squares  of  the  semi-axes  ( — j — ).  If  then 


CHAP.  VI.]  MOMENT   OF    INEKTIA.  71 

we  find  Jc  I,  the  mean  proportional  of  O  k  and  Jc  a,  and  lay  it  off 
twice  from  O  to  D  along  the  assymptote  O  K,  O  D  is  the  diag- 
onal of  a  rectangle  whose  sides  are  the  principal  axes.  We 
thus  find  the  vertices  A,  A,  B,  B. 

We  can  thus  construct  the  curves.  Then  for  any  position 
of  the  axis  X  X  as  it  revolves  about  O,  we  can  find  the  cor- 
responding radius  of  gyration  and  consequently  the  moment  of 
inertia,  by  simply  drawing  tangents  to  the  curve  above  and 
below  the  new  position  of  X  X  w&&  parallel  to  it.  The  radius 
of  gyration  thus  obtained  measured  to  the  scale  of  length  and 
multiplied  by  the  algebraic  sum  of  the  forces,  or  0  4  to  the 
scale  of  force,  will  give  the  moment  of  inertia  required  for  the 
assumed  position  of  the  axis. 

59.  Central  Curve.  Central  Ellipse. — If  the  point  O  about 
which  the  axis  turns  coincides  with  the  centre  of  action  (or 
gravity)  of  the  forces,  we  call  the  curve  enclosed  by  the  paral- 
lels M'  M"  at  the  distance  k  on  either  side,  the  central  curve. 
When  the  parallel  forces  all  act  in  the  same  direction  this  curve 
is  always  an  ellipse. 

For  the  central  curve  the  principle  proved  in  Art.  57  and 
the  method  of  construction  given  in  Art.  58,  are  no  longer 
applicable,  for  the  algebraic  sum  of  the  statical  moments  of 
the  given  forces  is  zero  for  every  axis  through  the  centre  of 
gravity.  We  cannot  therefore  find  the  centre  of  gravity  of  the 
moments  of  the  forces,  wrhen  considered  as  forces  themselves 
and  applied  at  the  given  points  of  application. 

If  we  divide,  however,  these  moments  considered  as  forces 
into  two  portions  or  groups,  and  find  the  centre  of  gravity  of 
each  group,  the  line  joining  these  two  points  has  an  important 
property,  viz.,  that  for  every  moment  axis  parallel  to  it,  the 
algebraic  sum  of  the  moments  of  the  statical  moments  consid- 
ered as  forces,  that  is,  the  algebraic  sum  of  the  moments  of 
inertia  of  the  forces,  is  zero.  In  other  words,  2  P  e  e'  is  zero, 
e  being  the  distances  of  the  points  of  application  from  the  first 
axis,  which  passes  through  the  centre  of  gravity  of  the  forces, 
and  e'  the  distances  from  the  axis  parallel  to  the  line  joining 
the  two  centres  of  gravity  of  the  two  groups  of  statical  moments 
considered  as  forces.  If  we  draw  then  through  the  centre  of 
gravity  of  the  forces  themselves  the  moment  axis  X  X,  and 
take  it  as  the  axis  of  abscissas  of  a  co-ordinate  system  whose  Y 
axis  passes  also  through  the  centre  of  gravity  of  the  forces  and 


72  MOMENT    OF   INERTIA.  [CHAP.  VI. 

is  parallel  to  the  line  joining  the  two  centres  of  gravity  of  the 
statical  moments  considered  as  forces,  then  the  moments  of 
inertia  2  P  y  x  are  zero,  and  hence  as  in  the  preceding  Art. 
this  axis  ofYis  conjugate  to  X  X. 

This  holds  good  not  only  for  the  central  curve,  but  also  for 
every  inertia  curve,  whose  centre  O  instead  of  coinciding  with 
the  centre  of  gravity  of  the  forces,  lies  in  the  axis  passing 
through  that  centre.  In  this  case  also  the  axis  through  the 
centre  O  parallel  to  the  line  of  union  above,  is  a  conjugate  to 
X  X.  Still  more,  the  half  length  of  this  conjugate  diameter  is 
in  both  cases  the  radius  of  gyration  of  the  force  system  for  the 
axis  X  X  and  the  direction  of  Y. 

Hence  in  every  inertia  curve  of  a  system  of  parallel  forces, 
whose  centre  lies  in  an  axis  passing  through  the  centre  of  grav- 
ity of  the  forces,  the  diameters  conjugate  to  this  axis  are  paral- 
lel and  equal.  All  these  inertia  curves  are  therefore  touched 
by  two  lines  parallel  to  this  axis  and  equally  distant  on.  either 
side.  This  distance  is  the  radius  of  gyration  for  this  axis. 

For  any  such  inertia  curve,  whose  centre  O  is  distant  i  from 
the  centre  of  gravity  S  of  the  forces,  we  call  E  and  (£  the  par- 
allel conjugate  axes  to  S  O  for  this  curve,  and  the  central  curve 
respectively;  q  and  q  the  distances  from  them  of  any  point  of 
application,  these  distances  measured  parallel  to  S  O,  and  con- 
sidered positive  when  the  point  of  application  lies  on  the  same 
side  of  E  or  (S  respectively  as  the  centre  of  gravity  S  from  E. 
Then  *,  the  distance  apart  of  E  and  (£  is  essentially  positive, 
and  if  we  indicate  by  a  and  a  the  lengths  of  the  semi-conj  ngate 
diameters  for  the  inertia  and  central  curve  respectively,  we 
have 


where  q  and  q  stand  in  the  simple  relation 

q  =  q+i. 
Hence 


Since  (§:  passes  through  the  centre  of  gravity  $P  q  =  o,  and 
therefore 

2  P  q2  =  5  P/  +«*  S  P  =  a*  2  P. 
Hence 

#  =  aa+£, 

an  equation  which  gives  the  relation  between  the  lengths  of 
the  semi-conjugate  diameters  of  the  central  and.  any  inertia 


CHAP.  VI.]  MOMENT   OF   INERTIA.  73 

curve,  whose  centre  lies  upon  an  axis  through  the  centre  of 
gravity  of  the  forces,  at  a  distance  \'  from  this  centre. 

Any  two  curves  at  equal  distances  either  side  of  the  centre  of 
gravity  are  therefore  equal.  If  the  semi-diameter  of  the  central 
curve  a  is  real,  and  therefore  a2  positive,  a2  is  also  positive  and 
greater  than  a2.  All  the  inertia  curves  are  therefore  of  the 
same  kind  as  the  central  curve,  and  enclose  the  centre  of  grav- 
ity. If,  however,  a2  is  negative,  and  the  central  curve  there- 
fore an  hyperbola;  all  those  inertia  curves  whose  centres  are 
distant  from  the  centre  of  gravity  by  a  distance  i  less  than  a 
are  hyperbolas  also.  For  a  distance  i  equal  to  a,  the  curves  re- 
duce to  straight  lines  equal  and  parallel  to  the  conjugate  diame- 
ter of  the  central  curve.  For  i  greater  than  a,  the  curves  be- 
come ellipses. 

6O.  Centre  of  Action  of  tlie  Statical  Moments  of  the 
Force§.*  —  We  again  suppose,  through  the  centre  of  gravity  of 
the  forces  S  [Fig.  35,  PL  11]  a  line  N  N  drawn  which  cuts  the 
central  curve  at  A  and  A'.  Two  such  points  we  have  in  every 
case,  except  when  the  curve  is  an  hyperbola,  andN  N  coincides 
with  an  assymptote. 

Let  (5  be  the  conjugate  axis  to  N  N  in  the  central  curve,  E  a 
parallel  to  it  through  any  point  o  distant  i  from  S,  and  also 
conjugate  to  N  N  in  the  inertia  curve  whose  centre  is  o.  Then 
since  the  statical  moments'  of  the  forces  with  reference  to  N  N 
is  zero,  the  centre  of  gravity  of  the  statical  moments  with  re- 
spect to  E?  considered  as  forces  acting  at  the  points  of  applica- 
tion, will  be  somewhere  upon  N  N.  It  is  required  to  find 
where. 

We  call  q  the  distance  of  any  point  of  application  from  E, 
measured  parallel  to  N  N,  and  positive  when  upon  the  same 
side  of  E  as  S,  then  i  is  essentially  positive. 

As  before,  q  is  the  distance  of  the  points  of  application  from 
(S,  also  measured  parallel  to  N  N,  and  positive  in  the  same 
direction  as  q. 

Then  we  have  always 

q  =  q+i. 

and  for  the  moments  of  inertia  of  the  forces  with  respect  to  E 
and(£  2P<?  =  ZP  (q+i)2  =  2Pf+#2P 

or  when  a  is  the  semi-diameter  of  the  central  curve,  S  A  =  S  A' 
and  ^Pq2=   a*+$  2P 


*  See  Supplement  to  Chap.  VIL,  Art.  10,  latter  part. 


74  MOMENT   OF   INERTIA.  [CHAP.  VI. 

Let  now  m  be  the  distance  of  the  centre  of  gravity  or  action, 
of  the  moments  of  the  force's  with  respect  to  E,  from  E,  and  in 
its  distance  from  (5,  positive  the  same  as  q  and  q.  Then 

m  =  tn  +  i 

and  since  the  sum  of  the  moments  is  equal  to  the  moment  of 
the  resultant  : 

^•Pq2  —  m^Pq. 

But  the  sum  of  the  moments  P  q  of  the  forces  with  reference 
to  E,  is  equal  to  the  product  of  the  sum  of  the  forces  into  the 
distance  i  of  the  centre  of  gravity  of  the  forces  from  E.  Hence 


and  therefore 

m  i  2  P  =  2P  q2  =  (a 
or,  m.  i  —  a2  +  r. 

Introducing  the  value  for  m 

(m  +  i)  i  =  a3  -f  ^2 
or  tit  i  —  a2. 

If  now  a2  is  positive,  which  is  always  the  case  for  an  ellipse 
as  central  curve,  m  is  also  positive,  and  is  therefore  to  be  laid 
off  from  S  along  N  N  on  the  opposite  side  of  Qj  from  o.  If 
then  we  conceive  an  axis  E'  drawn  parallel  to  E,  and  symmet- 
rical with  reference  to  S,  which  axis  we  shall  call  for  conven- 
ience the  symmetrical  axis  to  E,  we  see  from  the  above  relation 
that  M  is  the  pole  of  this  axis  in  the  central  curve. 

If,  however,  a2  is  negative,  therefore  a  imaginary,  nt  is  nega- 
tive, and  must  be  laid  off  from  S  towards  0,  and  the  point  M 
thus  found  is  therefore  the  pole  of  the  axis  E  itself,  or  in  the 
case  of  an  hyperbola  is  the  pole  of  E'  in  that  hyperbola  which 
is  not  cut  by  N  N,  and  for  which  therefore  A  A!  is  imaginary. 

Hence  we  have  the  principle  — 

If  we  consider  the  statical  moments  of  the  forces  with  refer- 
ence to  any  axis  as  E  as  themselves  forces  acting  at  the  given 
points  of  application,  the  centre  of  gravity  of  these  moment 
forces  does  not  coincide  with  the  centre  of  gravity  of  the  origi- 
nal forces,  but  is  the  pole  *  in  the  central  curve  of  an  axis  E' 
parallel  and  symmetrical  to  E. 

In  those  cases  where  the  central  curve  becomes  an  hyper- 

*  POLAR  LINE  OF  A  POINT,  in  the  plane  of  a  conic  section,  is  a  line  such, 
that  if  from  any  point  of  it  two  straight  lines  be  drawn  tangent  to  the  conic 
section,  the  straight  line  joining  the  points  of  contact  will  pass  through  the 
given  point,  which  is  called  a  pole. 


CHAP.  VI.]  MOMENT   OF   INERTIA.  75 

bola,  we  must  observe  whether  the  diameter  conjugate  to  the 
moment  axis  is  real  or  imaginary.  In  either  case  the  centre  of 
gravity  is  the  pole  of  the  line  symmetrical  to  the  moment  axis 
in  that  hyperbola  for  which  that  diameter  is  real  or  imaginary. 

The  construction  is  given  in  PL  11,  Fig.  35. 

Upon  S  o'  =  S  o  we  describe  a  semi-circle.  With  S  as  cen- 
tre, and  S  A'  =  a  =  semi-diameter  of  the  central  curve,  describe 
an  arc,  and  from  the  intersection  with  the  semi-circle  drop  a 
perpendicular  upon  So'.  The  point  M  thus  found  is  the  centre 
of  gravity  of  the  moments.  For  :  a2  =  a  M*  +  m2  and  a  M2 
=  ra  (i—m)  hence  a2  =  m2+m  i— m2  —  m  i.  The  central  curve 
being  known  as  also  the  distance  i,  the  point  M  can  be  readily 
found. 

61.  Cases  where  the  Direction  of  the  Conjugate  Axis  of 
the  Inertia  Curve  can  be  at  once  Determined. — There  are 
certain  special  and  practical  cases  in  which  the  conjugate  direc- 
tions or  axis  of  the  inertia  curve  can  be  at  sight  determined,  so 
that  only  the  length  of  the  semi-diameters  remains  to  be  found. 
The  most  important  of  such  cases  are  as  follows : 

(1.)  When  in  a  system  of  parallel  forces,  these  forces  can  be 
so  grouped  in  pairs,  that  the  lines  joining  the  points  of  appli- 
cation of  each  pair  are  all  parallel,  and  the  centres  of  gravity 
of  each  pair  all  lie  in  the  same  straight  line.  Then  for  the 
central  curve  and  all  inertia  curves  whose  centres  lie  upon  this 
straight  line,  the  direction  of  the  axis  conjugate  to  this  line  is 
the  same  as  that  of  the  lines  joining  the  points  of  application 
of  each  pair. 

This  is  easy  to  prove.  For,  for  each  pair,  the  sum  of  the 
moments  with  respect  to  the  line  joining  their  centres  of  gravity, 
is  zero.  These  moments  regarded  as  forces  and  applied  at  the 
points  of  application,  give  therefore  for  each  pair  two  parallel 
opposite  and  equal  forces,  the  sum  of  the  moments  of  which 
for  any  line  parallel  to  the  line  joining  the  points  of  applica- 
tion, is  zero.  This  is  the  case  for  all  the  pairs,  and  therefore 
the  direction  of  the  lines  joining  the  points  of  application  is 
that  of  the  axis  conjugate  to  the  line  joining  the  centres  of 
gravity,  for  the  central  curve  as  also  all  inertia  curves  whose 
centres  lie  upon  this  last  line. 

(2.)  When  the  forces  can  be  so  grouped  that  the  points  of  ap- 
plication of  each  group  lie  in  parallel  lines,  and  the  centres  of 
gravity  of  the  groups  lie  in  the  same  straight  line.  Then  this 


76  MOMENT   OF   INEBTIA.  [CHAP.  VI. 

straight  line  gives  the  direction  for  the  central  curve  and  every 
inertia  curve  whose  centre  lies  upon  it,  of  the  diameter  conju- 
gate to  an  axis  passing  through  the  centre  and  parallel  to  the 
lines  joining  the  points  of  application. 

For  if  we  take  any  such  axis,  the  points  of  application  of  the 
forces  in  each  group  are  equally  distant.  The  statical  moments 
for  each  group  are  then  proportional  to  these  distances.  If, 
therefore,  they  are  considered  as  forces,  their  centre  of  gravity 
coincides  with  that  of  the  forces  themselves,  and  lies  therefore 
in  the  line  joining  the  centres  of  gravity  of  the  groups.  The 
centre  of  gravity  of  the  whole  force  system  lies  then  in  this 
line,  which  is  therefore  the  direction  of  the  axis  conjugate  to 
the  line  parallel  to  the  lines  joining  the  points  of  application, 
in  the  central  curve,  and  also  all  curves  whose  centres  lie  upon 
this  line. 

(3.)  When  the  forces  can  be  so  grouped  that  the  centres  of  the 
central  curves  of  each  group  lie  in  the  same  straight  line,  and 
the  diameters  in  each  curve  conjugate  to  this  line,  are  parallel. 
Then  in  the  central  curve  of  the  entire  system,  the  diameter 
conjugate  to  this  line  is  also  parallel  to  these  diameters.  For, 
for  any  axis  parallel  to  these  diameters,  the  centres  of  gravity 
of  the  moments  of  the  forces  in  each  group  lie  upon  the  line 
joining  the  centres  of  the  curves.  The  centre  of  gravity  of  the 
moments  for  the  entire  system  lies  then  also  upon  this  line, 
which  is  therefore  the  direction  of  the  axis  conjugate  to  an  axis 
parallel  to  the  diameters  of  the  curves,  for  any  inertia  curve 
whose  centre  lies  upon  this  line. 

In  all  these  cases,  if  the  directions  thus  found  are  perpendicu- 
lar, we  have  to  do  with  the  principal  axes. 

62.  Practical  Applications. — We  can  now  apply  the  above 
principles  to  practical  cases,  and  as  in  the  determination  of  the 
moment  of  inertia  of  irregular  figures,  we  have  to  deal  with 
triangles,  parallelograms  and  trapezoids,  we  have  first  to  con- 
sider these  three  cases. 

1st.  The  Parallelogram.    PL  11,  Fig.  36. 

The  moment  of  inertia  of  a  parallelogram  is,  as  is  well  known, 

^  ==  12  a  ^**    a  ke*ng  tne  breadth  and  5  the  depth. 


ax2  dx  =  —  ab* 


CHAP.  VI.]  MOMENT   OF   INERTIA.  77 


Hence  T&  =  —  b2  =  radius  of  gyration,  or  k=  *J -bx  —  b. 

That  is,  the  radius  of  gyration  is  a  mean  proportional  between 

1  ,        .  1  _ 
-b  and  -b. 

The  centre  of  gravity  of  the  parallelogram  is  at  O  the  inter- 
section of  the  diagonals,  and  this  is  therefore  the  centre  of  the 
central  curve. 

If  we  suppose  the  parallelogram  divided  into  laminae  parallel 
to  D  C,  and  suppose  each  lamina  divided  by  G  H  parallel  to 
B  C,  the  centres  of  gravity  of  each  will  lie  upon  G  H.  Right 
and  left  of  G  H  we  then  have  a  group  of  forces  whose  points  of 
application  lie  in  lines  parallel  to  G  H,  and  the  lines  joining 
any  pair,  one  on  each  side  of  G  H,  are  parallel.  By  (1)  of  the 
preceding  Art.,  therefore,  GH  and  EF  are  conjugate  axes  of 
the  central  curve.  For  the  lengths  of  the  half  diameters,  we 

find  the  mean  proportional  between  -^  b  and  -^  5,  -^  a  and  -^  #, 

respectively,  by  the  half  circles  B  F  and  BH.  We  thus  find  k 
and  &',  and  can  then  construct  the  central  ellipse  directly,  or 
find  the  principal  axes,  and  then  construct  it.  The  centre  of 
gravity  of  the  moments  of  the  parallelogram,  with  reference  to 
any  axis  parallel  to  A  B,  is  as  we  have  seen,  Art.  60,  the  pole  of 
a  line  parallel  and  equally  distant  from  O  on  the  other  side.  If 
we  draw  this  line  then,  as  D  C,  then  from  G  draw  two  tangents 
to  the  central  ellipse,  and  unite  the  points  of  tangency  by  a 
line ;  the  intersection  of  this  line  with  O  G  is  the  centre  of 
gravity  of  the  moments  of  the  forces  themselves  considered  as 
forces,  or  area  of  the  parallelogram,  with  reference  to  A  B. 

Zd.   Triangle.     PL  11,  Fig.  37. 

The  moment  of  inertia  of  a  triangle  for  the  axis  B  C  is 

.H5  a  A3  *  whence  #  =  -~  A2,  and  for  an  axis  E  F  distant  i  = 

L£  O 

•o  A,  which  passes  through  £he  centre  of  gravity, 
a*  =  tf-#  =       tf-     (Art.  59.) 


I 

a  '±=f  tfdx  =  -^-ah3,h  being  the  line  A  D,  a  =  B  O. 


78  MOMENT   OF   IKERTIA.  [CHAP.  VI. 

The  conjugate  axes  of  the  central  curve  are  by  principle  1  or 
2  of  the  preceding  Art.  E  F  and  A  D. 

The  above  value  of  a  is  then  the  length  of  the  semi- diameter 


along  A  D,  or  a  —  V  -^  h  x  -=  h.      That  is,  a  is  a  mean  propor- 

tional between  -^  h  and  -~  h.     This  is  found  by  the  semi-circle 

ODFig.  37. 

The  moment  of  inertia  of  the  triangle  with  respect  to  A  D  is 


•g  h  (~o^)3-      The   radius  of  gyration  then   is  y  -^  {-^af  = 

'  "2  \%a)x  ^  \va)  or  a  mean  proportional  between  —,  and 
|offorDC. 

This  is  given  by  the  semi-circle  on  D  G  =  ^  D  C,  and  we 

thus  have  the  four  points  1  2  3  4  of  the  central  ellipse,  and  the 
semi-diameters  0  1  and  0  3,  and  can  therefore  construct  it. 
From  the  central  ellipse  as  before,  we  can  find  the  centre  of 
gravity  of  the  moments  considered  as  forces  for  any  axis  par- 
allel to  B  C  or  A  D,  as  also  in  either  case,  the  radius  of  gyration 
and  therefore  moment  of  inertia,  for  any  axis  passing  through  O. 

3d.  Trapezoid.    PL  11,  Fig.  38. 

Here  the  lines  E  F  joining  the  centres  of  the  parallel  sides, 
and  G  H  parallel  to  these  sides,  aird  passing  through  the  centre 
of  gravity  0,  are  the  conjugate  axes  of  the  central  ellipse. 

For  the  axis  A  B  and  direction  E  F,  the  moment  of  inertia  is 


a  and  5  being  A  B  and  C  D,  and  h  =  E  F.     The  square  of 
radius  of  gyration  is  then 


1   .      ,.  ,      ~6  a+b 

2 


- 


CHAP.  VI.]  MOMENT   OF   INEKTIA.  79 

For  the  radius  of  gyration  for  G-  H,  at  a  distance  i  =  -r-  h 


-  we  nave 


+    H 

9 


la  +  Sb          I   ,2^  +  2^\2 

0?  =  %?  —  $  =-£•  -  —  T-  A2  —   rr  A2  I  -  --T-  1   = 

6    a  +  b          9       V  a  +  b  / 

1  /a 
J  *'• 

This  radius  a  is  half  the  diameter  along  E  F. 
To  construct  it,  put  (3  a)2  =  -^  A2  +  -/  —  T~T\2  ^2- 

Describe  a  semi-circle  upon  E  F,  and  at  the  centre  01?  arid 
at  the  intersection  of  the  diagonals  &,  erect  perpendiculars 


Ol  J  and  K  L.     Then  F"!2  =     A2  and  iTL2  =  7-^  A2,  since 

^ 

E  K  =  --  j  h  and  K  F  =  ---  7  A 
a  +  b  a  +  b 

K  L  equal  to  J  M  from  J,  we  have 


E  K  =  --  j  h  and  K  F  =  ---  7  A.     If  therefore  we  lay  off 
a  +  b  a  +  b 


and  hence  the  half  diameter  sought  is  one-third  F  M.     "We 
thus  find  0  1  and  0  2. 

To  find  the  other  semi-diameter  we  have  the  moment  of  in- 

ertia for  E  F   and   direction  G  H,  jg-  (a*  -f-  a*  I  +  a 
hence  the  square  of  the  radius  of  gyration  is 


-^ 
This  last  expression  is  easily  constructed^   In  the  right-angled 


triangle  F  B  N,  the  hypothenuse  F  N  =  A/  (-  a\2  +  (^  b\*> 
B  N  being  made  equal  to  C  E.     If  we  describe  then  a  semi- 

J       /  '  7iy*dy+     I  *h  a-^-y^y 


80  MOMENT   OF   INERTIA.  [CHAP.  VI. 

circle  upon  F  U  =  -~  F  N,  and  make  F  W  —  ^  F  N,  F  V  is  the 

semi-diameter  sought.  We  thus  find  0  3  and  0  4,  and  can  now 
construct  the  central  ellipse.  This  being  constructed  we  can 
find  the  centre  of  gravity  of  the  moments  with  reference  to  any 
axis  parallel  to  A  B  or  E  F,  according  to  Art.  60,  or  the  moment 
of  inertia  for  any  axis  through  0,  by  drawing  a  parallel  tangent 
to  the  ellipse.  The  distance  from  0  to  the  point  o£  tangency 
gives  then  the  radius  of  gyration  for  that  axis. 

Uh.  Segment  of  Parabola.    PL  11,  Fig.  39. 

Let  the  segment  be  limited  by  B  C  =  2  A,  and  A  D  =  I. 
Then  it  is  evident  that  these  two  axes  are  conjugate  (Art.  61), 
and  the  centre  of  the  central  curve  is  0,  the  ratio  of  A  0  to 
OD  being  as  3  to  2.  Hence  AD  and  E'F',  parallel  to  CD 
through  0,  are  conjugate  axes  of  the  central  curve.  To  find  the 
length  of  the  semi-diameters  along  these  axes  we  find  first  the 
moment  of  inertia  of  the  segment  with  reference  to  an  axis  Y  Y 
parallel  to  E'  F'  and  tangent  to  the  parabola  at  A.  We  have 
then  for  this  moment  of  inertia 


/' 

Jo 


where p  is  the  parameter  of  the  parabola,  and  I  —  A  D.     Since 

4 
3 


4 
the  area  of  the  segment  is  -~  h  I,  we  have  for  the  square  of  the 


radius  of  gyration 


The  square  of  the  radius  of  gyration  then  for  E'  F'  whose 

distance  from  A  is  i  =  —  I  is 
o . 


a  being  the  semi-diameter  along  A  D.  It  is  easier  here  to  com- 
pute a,  viz.,  a  =  0.26186  I,  and  lay  it  off  from  O,  thus  finding 
3  and  4. 

For  the  other,  semi-diameter  we  find  the  moment  of  inertia 
for  A  D  and  the  direction  E'  F'.     Thus 


CHAP.  VI.]  MOMENT   OF   INERTIA.  81 

The  radius  of  gyration  squared  is,  therefore,  • 


and  hence  the  radius  of  gyration  is  &  =  0.44721  A.  Laying 
this  off  from  0,  we  obtain  1  and  2,  and  can  therefore  now  draw 
the  central  ellipse. 

63.  Compound  or  Irregular  Cro§§-Section§. — Every  cross- 
section  may  be  divided  up  into  trapezoids,  triangles,  parallelo- 
grams and  parabolic  segments,  and  the  above  cases  will  aid  us, 
therefore,  in  the  application  of  the  graphic  method  to  compound 
or  irregular  cross-sections.  The  engineer  is  often  called  upon 
to  determine  the  moment  of  inertia  of  such  sections  as  the  T, 
double  T,  or  different  combinations  of  these  in  proportioning 
the  different  pieces  of  bridges,  such  as  chords,  struts,  floor-beams, 
etc.,  as  also  in  many  other  constructions.  The  calculation  for 
such  cross-sections  is  sometimes  very  laborious.  As  an  example 
'  of  the  application  of  the  graphical  method  best  illustrating  the 
above  principles,  we  take  the  cross-section  shown  in  Fig.  40, 
PL  12. 

First  we  divide  the  cross-section  into  a  series  of  trapezoids. 
The  first  segment,  bounded  by  a  curve,  we  may  consider  a  para- 
bolic area.  These  trapezoids  we  reduce  to  equivalent  rectangles 
of  common  base  a  [Art.  32],  and  take  the  corresponding  heights 
as  forces.  These  forces  we  lay  off  in  the  force  polygon  and 
choose  a  pole  C  at  distance  H  from  force  line,  drawing  CO,  01, 
C  2,  etc.  Parallel  to  these  lines  we  have  the  first  equilibrium 

polygon  I  II  III VIII,  the  intersection  of  the  two  outer 

sides  of  which  gives  the  point  of  application  of  the  resultant. 
The  intersection  S  of  the  resultant  with  the  axis  of  symmetry 
gives  the  centre  of  gravity  of  the  cross-section  [Art.  30].  The 
segments  o  1',  1'2',  2'3',  etc.,  cut  off  from  o  S,  give  the  statical 
moments  of  the  forces  with  reference  to  o  S  to  the  basis  H. 
We  now  choose  another  pole  C'  at  distance  H',  and  form  another 
force  polygon,  considering  these  moments  as  forces,  and  applied 
at  the  centres  of  action  of  the  moments  of  the  separate  areas 
into  which  the  whole  cross- section  has  been  divided.  These 
centres  of  action  can  be  determined  by  forming  the  central 
curve  for  each  area  according  to  Art.  62,  and  then  applying  the 
6 


82  MOMENT   OF   INERTIA.  [CHAP.  VI. 

principle  of  Art.  60.  A  little  consideration  will  show  that  these 
centres  of  gravity  will  coincide  approximately  with  the  centres 
of  gravity  of  the  areas  themselves,  except  for  areas  (3)  (4)  (5) 
and  (6).  Finding  then  for  these  areas  the  centres  of  action  of 
the  moments  considered  as  forces,  we  construct  the  equilibrium 
polygon  O'  I'  IT. . .  .VIII'.  The  distance  0"  8"  cut  off  by  the 
first  and  last  sides  of  this  polygon  gives  the  moment  of  inertia 
to  the  pole  distances  H  and  H'  and  the  reduction  base  a.  Thus 
0"  8"  measured  to  scale  of  force  and  multiplied  by  a  H  H'  is 
the  moment  of  inertia  of  the  cross-section  with  reference  to  o  S. 

In  TI  H'   (V'8" 

The  radius  of  gyration  is  then  k=\J = . 

a  0  8 

The  division  will  be  performed  if  we  take  H'  =  0  8  =  2  P. 
This  we  can  easily  do  now  without  drawing  a  new  polygon, 
since  what  is  required  is  the  intersection  of  the  outer  sides  only. 
Thus  take  a  new  pole  C/  distant  from  o  S,  H'  =  0  8.  Now  we 
know  that  each  side  of  the  new  polygon  for  this  pole  distance 
will  intersect  the  corresponding  side  of  the  first  in  a  line  paral- 
lel to  o  GI  [Art.  27].  Since  the  new  polygon  may  start  from 
any  point,  we  may  take  the  first  side  to  coincide  with  O  VIII'. 
Then  the  line  of  intersection  of  any  two  sides  is  O  VnT  8". 
Produce  any  side  as  IV'  V7  to  intersection  e  with  this  line  ;  from 
e  draw  e  a/  parallel  to  C/  4'. 

Through  a!  the  intersection  of  o'  I'  and  V  IV,  the  resultant 
of  (1)  (2)  (3)  and  (4),  must  pass.  The  change  of  pole  cannot 
affect  this  resultant,  which  must  therefore  pass  through  a/,  the 
intersection  of  <?  a/  with  the  vertical  through  a'  parallel  to  o  S. 
Hence  0/  at'  is  the  direction  of  the  last  side  of  the  new  poly- 
gon, and  Sf/0i"  is  the  moment  of  inertia  for  the  new  pole  dis- 
tance o  C/  =  0  8.  The  radius  of  gyration  then  is  k  =  V  H  O/'S"! 
In  other  words,  Jc  is  a  mean  proportional  between  H  and  O/'S". 
The  construction  of  k  is  given  by  the  semi-circle  described  upon 
O/'S"  +  H.  The  ordinate  to  this  semi-circle  through  O/7  per- 
pendicular to  <?S  gives  Jc.  We  thus  find  the  semi-diameter 
S  a  =  S  a'  of  the  central  ellipse. 

In  order  to  find  the  other  semi-diameter  S  b  =  S  £',  we  might 
divide  the  cross-section  into  areas  by  lines  parallel  to  S  X,  and 
then  proceed  as  above.  This  is,  however,  unnecessary.  "With 
the  same  areas  as  before,  we  can  find  the  central  curve  for  that 
area  on  each  side  of  X  X,  and  then  the  centre  of  application  of 


CHAP.  VI.]  MOMENT   OF   INERTIA.  83 

the  moment  of  each  of  these  areas  with  respect  to  X  X  itself, 
considered  as  a  force.  The  method  of  procedure  is  then  pre- 
cisely as  before.  We  draw  a  polygon  the  sides  of  which  are 
respectively  perpendicular  to  those  of  the  first  polygon,  and 
thus  find  the  statical  moments  0"'  I'"  V"  2'",  etc.,  to  basis  H. 

Choosing  then  a  pole  C'"  at  distance  H'"  and  drawing  the 
corresponding  polygon,  we  have  0  8  1V  for  the  moment  of  in- 

ertia.    The  radius  of  gyration  is  then  Jc  =  \  aHH"'  Q8IV 

a.  0  8 


We  have  taken  H'"  =  g  08,  hence  k  =  1        H  QQ-  iv     Hence  k 

1  _ 

is  a  mean  proportional  between  g  0  8  1V  and  H.     The  construc- 

1  _ 

tion  is  given  in  the  Fig.  by  a  semi-circle  upon  H+  ^  0  8  IV.   We 

thus  find  the  semi-axis  S  V  —  S  £,  and  can  now  construct  the 
central  ellipse.  We  have  thus  found  graphically  not  only  the 
moments  of  inertia  of  the  cross-section  with  respect  to  X  X  and 
Y  Y,  but,  by  means  of  the  central  ellipse,  for  any  other  axis  in 
the  plane  of  the  Fig.  passing  through  S. 

64.  —  The  above  method  of  procedure  holds  good  generally 
for  any  cross-section,  except  that,  when  there  is  no  axis  of  sym- 
metry, the  centre  of  gravity  must  be  found  by  a  second  equili- 
brium polygon  whose  sides  are  respectively  perpendicular  to 
those  of  the  first.  When  the  moment  of  inertia  with  reference 
to  a  single  axis  only  is  required,  the  above  method  becomes 
quite  short  and  simple,  as  well  as  accurate.  In  our  Fig.  the 
scale  used  as  also  the  number  of  divisions  taken  make  the  pro- 
cess appear  more  complicated  than  it  really  is. 

With  this  we  shall  close  our  discussion  of  moment  of  inertia, 
merely  observing,  that  all  the  principles  deduced  in  this  chap- 
ter for  forces  acting  in  a  plane  hold  equally  good  for  forces  in 
space.  The  central  curve  then  becomes  an  area,  we  have  a  mo- 
ment plane  instead  of  moment  axis  M,  and  the  ellipse  and  hyper- 
bola of  inertia  become  ellipsoid  and  hyperboloid  respectively*. 

For  a  much  fuller  discussion  of  the  subject  than  is  possible 
here,  we  refer  the  reader  to  Culmanrfs  Graphische  Statik,  pp. 
160-206  ;  also  Bauschinger's  Elemente  der  Graphischen  Statik, 
pp.  116-168.  To  the  latter  we  are  largely  indebted  in  the 
preparation  of  the  present  chapter  ;  Plates  10  and  12  are,  with 
slight  alteration,  reproduced  from  that  work. 


84-  APPLICATION  TO  BRIDGES.  PART -II. 


PART    II. 

APPLICATION  TO  BRIDGES. 


65.  —  Under  the  head  of  Parallel  Forces  we  have  already 
given  the  general  application  of  the  graphical  method  to  the 
determination  of  the  moments  and  shearing  forces  in  beams 
resting  upon  two  supports  only.     We  shall  now  take  the  sub- 
ject up  more  in  detail,  and  show  the  methods  of  determining 
the  maximum  strains  for  all  the  possible  conditions  of  loading 
which  may  occur  in  Bridge  Girders.     In  the  following  we  shall 
adhere  closely  to  the  development  of  the  subject  as  given  by 
Winkler.     {DerBruckenbau,  Wien,  1872.] 

66.  Force§  wliicli  act  upon  a  Bridge.  —  The  forces  which 
act  upon  a  bridge  may  be  enumerated  as  follows  : 

1st.  The  weight  of  the  bridge  itself.  —  This,  previous  to  the 
calculation  of  the  strains,  is  unknown,  since  it  depends  upon  the 
intensity  of  the  strains  themselves.  It  is  customary  to  assume 
the  weight  to  begin  with,  by  comparison  with  existing  struc- 
tures of  similar  character,  and  then  to  find  the  resulting  strains. 
The  weight  answering  to  these  strains  can  then  be  easily  ascer- 
tained ;  the  strength  of  the  materials  used  being  known,  and 
compared  with  the  assumed  weight.  According  as  it  is  less  or 
greater,  the  weight  was  then  assumed  too  great  or  the  reverse, 
A  second  approximation  to  the  true  weight  may  then  be  made, 
and  the  strains  proportionally  diminished  or  increased.  As 
rules  for  estimating  the  weight  of  bridge  girders  under  200  feet 
span,  we  have,  for  weight  of  girder  G-, 


where  W  =  the  assumed  approximate  total  distributed  load  in 
tons,  including  the  weight  of  girder  ; 

I  =  length  in  feet  ; 

d  =  depth  in  feet  ; 


PAKT   II.]  APPLICATION   TO   BRIDGES.  85 

y=  the  working  strain  in  tons  per  sq.  foot  of  cross-section. 
(See  Stoney,  Theory  of  Strains,  vol.  ii.,  p.  441.) 

We  have  also  the  rule :  "  Multiply  the  distributed  load  in 
tons  by  4 ;  the  product  is  the  weight  of  the  main  girders,  end- 
pillars  and  cross-bracing  in  pounds  per  running  foot."  Iron  is 
taken  at  5  tons  per  sq.  inch  tension,  and  4  tons  per  sq.  inch 
compression. 

2d.  The  moving  or  live  load;  which  is  determined  by  the 
purpose  of  the  bridge.  This  load  can  take  various  positions 
upon  the  bridge,  and  may  even  be  divided  into  several  por- 
tions. It  is  therefore  an  important  problem  to  determine  that 
distribution  which  shall  cause  the  maximum  strains. 

The  live  load  is,  as  the  term  implies,  in  motion,  so  that,  in 
combination  with  the  deflection,  there  is  a  centrifugal  force, 
or  increase  of  pressure.  This  is,  however,  in  practice  disre- 
garded, while  such  a  coefficient  of  safety  is  chosen  in  propor- 
tioning the  parts,  that  the  increase  of  strain  due  to  this  cause  is 
fully  covered. 

3d.  Horizontal  forces,  caused  by  the  wind  and  the  passage 
of  loads. 

^th.  Pressures  at  the  supports.  The  known  forces  cause  re- 
actions at  the  supports,  which  evidently  must  also  be  considered 
as  forces  acting  upon  the  bridge  girder.  For  straight  girders, 
these  reactions  are  vertical,  while  in  suspension  and  arch  sys- 
tems they  are  inclined. 

67.  Bridge  Loading. — The  heaviest  load  to  which  a  railway 
bridge  can  be  subjected  is  when  it  is  covered  from  end  to  end 
with  locomotives.     "  The  standard  locomotive  is  assumed  to  be 
24  feet  long,  and  to  have  six  wheels  with  a  12-foot  base ;  to 
have  half  its  weight  resting  on  the  middle  wheels,  and  one- 
fourth  on  the  leading  and  trailing  pairs  respectively,  which  are 
supposed  to  be  at  equal  distances  on  either  side  of  the  middle 
wheels."     (See  Stoney,  vol.  ii.,  p.  405.)     The  standard  engine 
is  assumed  to  weigh  24  tons.  30  tons  and  32  tons,  according  to 
the  construction.     This  makes  the  standard  load  1  ton,  1J  ton,  or 
1^  ton  per  foot  of  single  line.     Short  bridges  of  less  than  40 
feet  span  must  be  considered  as  subject  to  concentrated  loads 
from  single  engines. 

The  maximum  load  for  public  bridges  is  recommended  by 
Stoney  at  100  Ibs.  per  sq.  ft. 

68.  In  the  Straight  Truss  all  the  Outer  Forces  act  in  a 


86  APPLICATION   TO   BRIDGES.  [PAET   If. 

Vertical  Direction. — The  strain  in  any  cross-section  depends 
upon,  first,  the  resultant  of  all  the  outer  forces  acting  either 
side  of  the  cross-section ;  and  second,  the  statical  moment  of 
these  forces  with  reference  to  the  cross-section.  The  first,  or 
the  algebraic  sum  of  all  the  forces  acting  between  the  cross- 
section  and  either  end,  we  call  the  shearing  force  for  this  cross- 
section,  and  indicate  it  by  S.  It  is  also  designated  as  vertical 
force,  or  transverse  force.  The  moment  of  the  resultant,  or 
the  algebraic  sum  of  the  moments  of  all  the  exterior  forces, 
with  reference  to  any  cross-section,  we  call  the  moment  for  this 
cross-section,  and  indicate  it  by  M.  It  is  also  called  bending 
moment,  or  moment  of  rupture.  For  example,  in  a  lattice 
girder  with  horizontal  flanges  the  strains  in  the  web  are  pro- 
portional to  the  shearing  forces,  those  in  the  flanges  to  the 
bending  moments. 

The  shearing  force  is  considered  positive  when  it  acts  on  the 
left  side  upwards,  or  on  the  right  side  downwards.  The  mo- 
ment M  is  positive,  when  on  the  left  side  the  tendency  of  rota- 
tion is  to  the  left,  on  the  right  side  to  the  right,  or  when  it  tends 
to  make  the  girder  convex  upwards,  that  is,  causes  compression 
in  the  lower  fibre  or  flange. 


CHAP.  VH.]  SIMPLE   GIKDEKS.  87 


CHAFTEE   VII. 


SIMPLE   GIKDEKS. 


69.  Action  of  Concentrated  Loads — Invariable  in  Posi- 
tion.— By  "simple  girder"  we  understand  a  girder  resting 
upon  two  supports  only,  in  opposition  to  a  continuous  girder 
which  rests  upon  more  than  two. 

Suppose  a  number  of  forces  Pl  .  .  .  P5  acting  at  various  points. 
[Fig.  41,  PI.  13.]  We  form  the  force  polygon  by  laying  off  the 
forces  to  scale  one  after  another ;  then  choose  a  pole  O,  and 
draw  O  0, 0 1, 0  2,  etc.,  to  the  points  of  division.  Parallel  to  these 
lines  we  draw  the  lines  of  the  equilibrium  polygon  between  the 
corresponding  force  lines  prolonged.  If  now  we  dose  the  poly- 
gon thus  formed  by  the  line  A  B,  and  draw  through  O  the 
parallel  O  L  to  A  B,  the  segments  0  L  and  L  5  of  the  force 
line  give  the  reactions  Vj_  and  V2.  Further,  the  shearing  force 
between  A  and  Pt  is  St  *=  V^  =  L  0  ;  between  Pl  and  P2,  S2  = 
V\— P!  ;  at  P3,  S3  =  Vt— P,,—  P2,  etc.  That  is,  the  shearing  forces 
are  the'  distances  of  the  points  of  the  force  polygon  fro  in~L>.  It 
is  easy,  then,  to  construct  them,  as  shown  in  the  lower  shaded 
area  of  the  Fig.  (See  also  Art.  46.) 

If  in  the  equilibrium  polygon  we  let  fall  at  any  point  a  ver- 
tical as  I  K,  and  from  K  draw  K  L  perpendicular  to  A  B,  and 
indicate  by  H  the  horizontal  pull,  by  L  the  strain  in  A  B,  and 
by  M  the  sum  of  the  moments  of  all  forces  left  of  I  K,  then, 
for  equilibrium  about  K,  we  have  M^LxKL^LxIK  cos 
I  K  L,  or,  since  the  angle  I  K  L  =  L  O  H  in  force  polygon, 
L  x  cos  I  K  L  =  H,  and  hence  M  =  H  x  I  K,  or  representing 
the  variable  ordinate  I K  by  y : 

M  =  Hy. 

But  H  is  the  distance  of  the  pole  O  from  the  force  line ; 
the  moment  at  any  point  is  therefore  proportional  to  the  verti- 
cal height  of  the  equilibrium  polygon.  (See  also  Art.  38.)  If 
we  take  H  equal  to  the  unit  of  force,  we  have 


88  SIMPLE    GIRDERS.  [CHAP.  VII. 

so  that  in  this  case  the  moment  at  any  point  is  directly  given 
by  the  ordinate  of  the  polygon  at  that  point.  It  is  this  impor- 
tant property  of  the  equilibrium  polygon  which  renders  it  espe- 
cially serviceable  in  the  graphical  solution  of  this  and  similar 
problems. 

70.  Concentrated  Load  —  Variable   Position— Shearing 
Force. — If  the  load  lies  to  the  right  of  any  given  cross-section, 
then  the  shearing  force  at  this  cross-section  will  be  S^V^  or, 
since  we  regard  a  force  to  the  left  acting  up  as  positive,  S  is 
positive.     As  the  load  P  moves  towards  the  left,  Vt  or  S  in- 
creases.    When  the  load  is  to  the  left  of  the  cross-section,  the 
shearing  force  at  the  cross-section  is  S  =  V^  —  P,  and  since  P 
is  always  greater  than  Vl5  S  is  negative.     The  nearer  P  ap- 
proaches the  cross-section,  the  smaller  is  S. 

Hence :  a  concentrated  load  causes  a  positive  or  negative 
shear,  according  as  it  is  to  the  right  or  left  of  the  cross-section 
considered,  and  the  shearing  force  is  greater  the  nearer  the  load 
is  to  the  cross- section. 

Moments. — If  the  load  lies  to  the  right  of  the  cross-section, 
the  moment  is  M  =  —  V\  x,  x  being  the  distance  of  the  cross- 
section  from  the  left  support.  M  is  therefore  negative  and  in- 
creases with  Y!  ;  that  is,  as  the  load  .approaches  the  cross-sec- 
tion. If  the  load  is  on  the  left  of  the  cross-section,  M  —  —  V2 
(I  —  #),  V2  being  the  reaction  at  the  right  support.  Here  also 
M  is  negative  and  increases  with  V2 ;  that  is,  as  the  load  ap- 
proaches the  cross-section. 

Hence :  a  concentrated  load  wherever  it  lies  causes  in  every 
cross-section  a  negative  moment,  which  for  any  cross-section  is 
a  maximum,  when  the  load  is  applied  at  that  cross-section. 

71.  Position  of  a  given  System  of  Concentrated  Loads 
causing  Maximum  Shearing  Force. — If  P!  is  the  sum  of  all 
the  loads  to  the  left  of  any  cross-section,  the  shear  at  that  cross- 
section  is  S  =  Vj  —  Px.     As  the  system  moves  to  the  left  with- 
out any  load  passing  off  the  girder  or  any  load  passing  the  cross- 
section,  Y!  and  therefore  S  increases  as  long  as  S  is  positive,  or 
as  long  as  Vx  >  Pl8     If  a  load  passes  off  the  girder,  then  for  the 
remaining  loads  S  increases  anew  as  the  system  moves  to  the 
left,  until  a  load  of  the  system  passes  the  cross-section  in  ques- 
tion.    The  same  holds  good  for  a  system  moving  to  the  right, 
where  S  is  negative. 

Hence :    the  shearing  force  is  a  maximum  for  any  point, 


CHAP.  VII.]  SIMPLE   GIRDERS.  89 

when  there  is  a  load  of  the  system  at  that  point,  and  the  maxi- 
mum is  positive  or  negative,  according  as  the  load  lies  just  to 
the  right  or  left  of  the  point. 

Since  for  a  single  load  (Art.  70)  S  is  positive  or  negative,  ac- 
cording as  the  load  is  to  the  right  or  left,  S  will  be  in  general 
a  positive  or  negative  maximum  when  all  the  loads  lie  to  the 
right  or  left,  and  the  heaviest  nearest  the  cross-section.  Only 
in  cases  where  a  small  load  precedes,  can  S  be  greatest  when 
the  second  load  lies  upon  the  point  in  question. 

If  P  is  the  resultant  of  all  the  loads  and  j3  its  distance  from 
the  right  support, 

Vt  =  P  £  and  therefore  S  =  P  £  —  ple 
L  L 

Now  the  position  of  the  loads  of  the  system  or  @  remaining 
unchanged,  Px  will  vary  as  the  first  power  of  x,  the  distance  of 
the  cross-section  from  the  left  support.  Therefore,  between  any 
two  cross-sections  for  which  the  load  on  the  girder  remains  the 
same,  the  shear  S  is  represented  by  the  ordinates  to  a  straight 
line. 

72.  Construction  of  the  Maximum  Shearing  Forces.— 
Construct  the  force  polygon  with  the  given  loads ;  choose  a 
pole  O  [PL  13,  Fig.  42  (a)~\  and  draw  the  corresponding  equi- 
librium polygon.  It  is  required  to  determine  the  shear  S  at  a 
crossrsection  distant  x  from  the  left  support,  under  the  suppo- 
sition that  the  first  load  PL  of  the  system,  moving  towards  the 
left,  acts  at  this  cross-section. 

Determine  upon  the  outer  side  P!  A  of  the  polygon  passing 
through  the  point  Pl5  a  point  A  distant  from  Plt  by  the  distance 
x,  and  then  find  the  point  B  upon  the  polygon  distant  from  A 
by  I,  the  length  of  span,  and  draw  A  B.  Parallel  to  A  B  draw 
O  L  in  the  force  polygon,  then  A  L  =  Vt  =  S,  the  shear  at  Pt. 
Drop  a  vertical  through  B  intersecting  P!  A  produced,  in  M ; 
then  the  triangles  O  A  L  and  Pl  M  B  are  similar,  and  there- 
fore S  =  A  L  =  B  M  — ,  when  a  is  the  pole  distance.  If  we 

if 

choose  a  =  I,  then  S  =  B  M. 

Hence :  the  maximum  shearing  forces  are  proportional  to 
the  vertical  segments  between  the  equilibrium  polygon  and  the 
prolongation  of  the  outer  side  taken  at  the  end  of  the  system, 
or  are  equal  to  these  segments  if  the  pole  distance  is  taken  equal 
to  the  span  ;  provided  that  the  last  load  is  at  the  cross-section. 


90  SIMPLE   GIKDEKS.  [CHAP.  VII. 

We  have,  therefore,  the  simple  construction  given  in  PL  13, 
Fig.  42  (b).  The  broken  lines  are  parallel  to  the  various  posi- 
tions of  A  B  for  corresponding  positions  of  Pt.  The  positive 
and  negative  values  of  S  equally  distant  from  the  right  and  left 
supports  are  equal,  so  that  it  is  only  necessary  to  construct  S 
for  one  value. 

If  the  second  load  is  to  be  at  the  cross-section,  and  if  e  is  the 
distance  between  the  first  and  second,  we  draw  first  a  line 

&   _-_.-L-    np 

whose   equation   is   2/  —  PI  — -, — ,  and  construct,  as   above,  a' 

I/ 

polygon,  for  which  the  second  load  lies  on  the  right  support  B, 
and  whose  second  side  (between  second  and  third  loads)  coin- 
cides with  the  above  line.  The  ordinates  to  this  line  above  the 
axis  of  abscissas  will  give  maximum  of  +  S. 

73.  Maximum  Moments. — Since,  according  to  Art.  70,  a 
concentrated  load  causes  a  negative  moment  at  any  point, 
wherever  it  may  lie,  we  must  have  evidently  loads  upon  both 
sides  of  any  point,  in  order  that  the  moment  may  be  a  maxi- 
mum. Since  a  single  load  causes  a  greater  moment  at  any 
point  the  nearer  it  lies  to  that  point,  the  greatest  load  must  lie 
nearest  the  cross-section  in  question.  The  method  of  loading, 
causing  maximum  moments,  can  be  best  determined  for  a  dis- 
tributed load  (not  necessarily  uniform).  In  this  case  the  equi- 
librium polygon  becomes  a  curve  [PL  13,  Fig.  43].  If  in  this 
curve  we  draw  A  B,  and  take  C  so  that  A  C  :  C  B  \;x  :  I  —  #, 
then  C  D  =  M  f  or  x.  Suppose  A  B  to  take  the  position  A!  B', 
the  horizontal  protection  of  C  C'  being  indefinitely  small, 
then  C'  D'  =  M  +  d  M.  In  order  now  that  M  may  be  a 
maximum,  C'  D'  must  be  equal  to  C  D  or  C  C'  parallel  to  D  D'. 
If  in  the  force  polygon  O  Al  is  parallel  to  A  A',  O  Bx  to  B  B'. 
and  O  Dt  to  D  D',  then  A^  ^  and  E>!  B1  are  the  loads  upon 
A  C  and  B  C. 

Draw  through  C  a  vertical,  and  through  A,  A',  B,  B',  paral- 
lels to  C  C7  or  D  D'  intersecting  this  vertical  in  E,  E',  F,  F'. 

Then  CE:CF::AC:BC;:a;:Z-tf, 

C  E'  :  C  F'  ; ;  A'  C'  :  B7  C'  : ;  x  :  I  -  x; 
therefore 

C  E  :  C  F  ; ;  C  E'  :  C  F',  or  C  E'  :  C  E  ; :  C  F'  :  C  F ; 
also  CE'—  CE:CF'-CF;:CE:CF, 

that  is,  E  E7  :  F  F'  \\x\l-  x. 

If  now  we  draw  through  A'  and  B'  parallels  to  C  C',  or  D  D'  to 


CHAP.  VII.]  SIMPLE   GIRDERS.  91 

intersections  H  and  I,  we  have  A  H  =  E  E',  I  B'  —  F  F'.  Since 
the  triangle  A  A'  H  is  similar  to  O  A!  ^  and  B  B' 1  to  O  B1  T>ly 
and  since  A7  H  =  B  I,  we  have 

A!  Dt :  B!  D!  ; ;  A  H  :  B'  I ; ;  E  E' :  F  F7 ; ;  x  :  l—x. 

Since  A1  Dx  equals  the  load  Pt  on  A  C,  and  B!  "D^  the  load  P2 
on  B  C,  we  have  Pl  :  P2 ; ;  x  :  l—x. 

The  same  will  hold  true  approximately  for  concentrated 
loads.  Hence,  in  order  that  the  moment  at  any  point  may  be 
a  maximum,  the  system  of  loads  must  have  such  a  position 
that  the  loads  either  side  of  this  point  are  to  each  other  as  the 
portions  into  which  the  span  is  divided. 

In  PI.  13,  Fig.  44,  let  C  D  give  the  moment  at  C.  If  the  line 
A  B  moves  so  that  the  horizontal  projections  of  A  C  and  B  C 
remain  equal  to  x  and  I— x,  then  as  long  as  the  ends  A  and  B 
move  on  the  same  straight  lines,  the  point  C  will  also  move  in  a 
straight  line.  The  point  C  describes,  therefore,  a  broken  line. 
The  verticals  between  this  line  and  the  polygon  correspond  to 
the  moments  for  various  positions  of  the  load  and  a  given  value 
of  x.  Evidently  the  greatest  ordinate  will  be  over  an  angle  of 
the  equilibrium  polygon  which  is  not  under  an  angle  of  the 
line  described  by  C— that  is,  for  M  maximum  a  load  must  lie 
upon  the  cross-section. 

For  any  cross-section,  then,  the  moment  is  a  maximum  when 
a  load  is  applied  at  this  cross-section.  Which  of  the  loads 
must  be  so  applied  is  determined  by  the  preceding  rule. 

74.  Contraction  of  Maximum  Moments.  —  After  the 
equilibrium  polygon  has  been  constructed,  in  order  to  find  M 
for  a  point  C  (PI.  13,  Fig.  45),  we  determine  two  points  P  and 
G  upon  the  polygon  which  are  distant  horizontally  from  the 
load  on  the  given. cross-section  corresponding  to  the  angle  E  by 
distances  A  C,  B  C.  Then  draw  F  G-,  and  the  vertical  K  E  is 
equal  to  M  when  the  pole  distance  is  unity.  We  make  C I  =  E  K. 
In  this  way  we  can  construct  the  moments  for  different  loads 
of  the  load  system  at  the  given  cross-section,  and  thus  determine 
that  position  of  the  load  which  gives  the  maximum  moment  at 
the  cross-section. 

Generally  when  K  E  =  y,  and  the  pole  distance  is  a,  we  have 
JML  =  ay.  The  pole  distance  a  is  measured  to  the  scale  of 
force,  and  then  y  is  given  by  the  scale  of  length.  The  unit  for 

M,  in  order  that  M  may  be  equal  to  y,  is  evidently  -th  part  of 


92  SIMPLE   GIRDERS.  [CHAP-  VIT- 

the  unit  of  length  (when  the  pole  distance  is  a  force  units),  or, 
what  is  the  same  thing,  one  unit  of  length  is  equal  to  a  moment 
units.  The  same  equilibrium  polygon  can  be  used  for  any 
number  of  girders  of  various  spans,  hence  the  method  is  of  very 
rapid  application. 

75.  Ab§olute  maximum  of  Moments. — Since  for  any  cross- 
section  M  is  a  maximum  when  a  load  lies  at  that  section,  a  load 
must  also  lie  upon  the  cross-section  for  which  M  is  an  absolute 
maximum.  , 

If  the  line  A  B  slides  upon  the  equilibrium  polygon,  altering 
its  length  so  that  its  horizontal  projection  is  constant  and  equal 
to  I,  it  will  envelop  a  portion  of  a  parabola  so  long  as  its  ends 
move  in  the  same  sides  of  the  polygon.  [PL  13,  Fig.  46.]  The 
curve  thus  produced  is  therefore  composed  of  portions  of  a 
parabola.  Let  the  ordinate  D  C  correspond  to  the  moment  at 
the  point  of  application  of  the  load  P.  DC  will  be  evidently 
greatest  when  A  B  is  tangent  to  .the  curve  at  C,  so  that  the 
maximum  of  the  moments  occurring  at  D  is  given  by  the  dis- 
tance C  D  between  the  polygon  and  curve  enveloped  by  A  B. 

Let  the  prolongation  of  the  sides  upon  which  A  B  slides  meet 
in  E,  and  F  G  be  the  tangent  to  the  parabola  at  the  point  H  in 
the  vertical  through  E,  so  that  F  H  =  H  G-,  and  let  I  be  the  in- 
tersection of  A  B  and  F  G.  Draw  through  A  a  parallel  to  E  B, 
intersecting  F  G  in  K.  Then  the  horizontal  projections  of  A  F 
and  A  K  are  equal,  since  those  of  E  F  and  E  G  are  equal. 

Since,  however,  the  projections  of  F  G  and  A  B  as  also  of 
A  F  and  G  B  are  equal,  A  K  must  be  equal  to  G  B.  Hence 
A I  =  B  I.  In  a  parabola  the  distances  of  the  three  diameters 
passing  through  two  points  and  the  point  of  intersection  of  the 
corresponding  tangents  are  equal,  hence  the  projections  of  H  I 
and  C  I  are  equal. 

The  middle  point  I  of  the  tangent  A  B  lies,  then,  half  way 
between  the  angle  D  vertically  below  the  point  of  tangency  and 
the  intersection  E  of  the  sides  upon  which  it  slides. 

Since  the  projection  of  A  B  is  I,  its  construction  is  easy. 
The  construction  must,  of  course,  be  repeated  for  each  angle,  in 
order  to  determine  that  for  which  M  is  an  absolute  maximum. 

The  above  principle  may,  then,  be  thus  expressed  :  The  mo- 
ment at  any  load  is  a  maximum,  when  this  load  and  the  result- 
ant of  all  the  loads  are  equally  distant  from  the  centre  of  the 
girder.  (See  also  Art.  48.) 


CHAP.  Vn.]  SIMPLE    GIRDERS.  93 

76.  —  In  Arts.  46  to  50  the  above  principles  have  been  already 
deduced  so  far  as  relates  to  the  moments  alone,  and  a  reference 
to  Art.  49  will  show  their  application  to  the  investigation  of  the 
effect  of  a  system  of  loads  moving  over  the  girder.     We  pass 
on,  therefore,  to 

CONTINUOUSLY   DISTRIBUTED   LOADING. 

Suppose  the  load  p  per  nnit  of  length  laid  off  as  ordinate. 
The  area  thus  obtained  we  call  the  load  area.  PI.  13,  Fig. 
46  {£). 

The  equilibrium  polygon  becomes  here  a  curve,  for  which 
the  same  law  holds  good.  If  we  draw  tangents  to  the  curve  at 
the  points  D/  and  E'  corresponding  to  D  and  E,  intersecting  in 
C',  then  the  resultant  of  the  load  upon  D  E  passes  vertically 
through  C',  or  C'  is  vertically  under  the  centre  of  gravity  of  the 
area  D  D"  E"  E. 

If  we  consider  the  load  area  divided  into  a  number  of  parts, 
the  resultant  for  each  will  pass  through  the  intersection  of  the 
tangents  at  the  points  vertically  under  the  lines  of  division. 
Since  these  tangents  are  parallel  to  the  lines  in  the  force  poly- 
gon corresponding  to  these  lines  of  division,  they  form  the 
equilibrium  polygon  for  the  concentrated  loads,  or  resultants  of 
the  portions  into  which  the  load  area  is  divided. 

Hence  :  if  we  divide  the  load  area  into  portions,  and  replace 
each  ~by  a  single  force,  the  sides  of  the  corresponding  polygon 
are  tangent  to  the  equilibrium  curve  at  the  points  correspond- 
ing to  the  lines  of  division.  (Art.  42.) 

77.  Total  Uniform  Load.  —  In  this  case  the  reactions  at  the 

supports  are  Vt  =  V2  =  —  p  I.    Hence,  for  any  cross-section  dis- 
tant x  from  the  left  support,  the  shearing  force  is 


For  a?  =  -~  Z  ;  S  =  0.     Sis  greatest  for  x  =  0  and  for  x  =  I  ; 

that  is,  maximum  S  =  -f  -p  Z,  and  S  =  —  ^  p  I. 
The  moment  at  any  cross-section  is 

M  =  —  V  x  +  -px*  =  —  -^px  (l—x). 

M  will  be  greatest  for  x  =  -=  l>  and 


94  SIMPLE   GIRDERS.  [CHAP.   VII. 


Max. 

s^ 

The  shearing  forces  are,  then,  given  by  a  straight  line  inter- 
secting the  span  in  the  middle,  the  ordinate  at  either  end  being 

Ip  I     [PI.  14,  Fig.  47.] 

The  moments,  as  we  have  already  seen  [Art.  44,  Fig.  30],  are 
given  by  a  parabola  whose  vertex  is  in  the  centre  of  the  span 

and  whose  middle  ordinate  is  -p  1?.    Since  we  have  seen  [Art. 

o 

70]  that  a  load  at  any  point  causes  at  every  point  a  negative 
moment,  the  maximum  moment  at  any  point  will  be  when  the 
whole  span  is  loaded. 

78.  Method  of  Loading  causing  Maximum  Shearing 
Force.  —  We  have  seen  [Art.  70]  that  a  single  load  causes  at 
any  point  a  positive  or  negative  shear,  according  as  it  lies  upon 
the  right  or  left  side  of  the  cross-section  at  that  point.  Hence, 
for  a  uniform  load, 

The  shearing  force  will  be  a  positive  or  negative  maximum 
according  as  the  load  reaches  from  the  right  or  left  support  to 
the  cross-section  in  question.  For  the  positive  maximum  we 

have  Y!  —p  (l—%)  ^r-y  —~^P  -  —  T~^-   Therefore,  max.  +  S  = 

a  L  2i  o 

1      (l-xf 


For  the  graphical  determination  we  can  apply  the  method 
given  in  Art.  72,  Fig.  42,  by  which  we  have  for  max.  +  S  and 
max.—  S  two  parabolas  whose  vertices  are  at  the  ends  of  the 

span,  and  whose  ordinates  at  these  points  are  +~r  and  —  •*-—• 

2i  2i 

Since,  however,  each  point  is  found  thus  from  the  preceding, 
the  construction  is  not  very  exact.  We  may  deduce  a  better 
construction  as  follows.  [PI.  14,  Fig.  48.]  Through  any  point 
F  of  the  curve  drop  a  vertical  intersecting  A  B  in  C  and  the 
line  B  K  parallel  to  the  tangent  at  F  in  G.  Let  the  tangent 

at  F  intersect  A  B  in  H.    Then  C  H  =  B  H  ;  hence,  C  F  =  -  C  G-. 

2 

We  have,  then,  A  E  =  i  A  D  =-1^  Z.     Since  C  F  =  -  C  G-, 

a  A  2 

we  have  also  AI  =-  .  AK  ;  therefore,  A  I  :  A  E  ;  ;  AK  :  A  D. 

a 


CHAP.  VII.]  SIMPLE    GIRDERS.  95 

Hence  the  following  construction  : 

Make  A  E  =  \.p  I. 

\ 
Divide  A  E  and  A  B  into  an  equal  number  of  equal  parts,  and 

draw  lines  from  B  to  the  points  of  division  of  A  E,  and  verticals 
through  the  points  of  division  of  A  B.  The  curve  passes 
through  the  points  of  intersection  of  corresponding  lines. 

79.  Live  and  Dead  Loads.  —  Let  p  be  the  load  per  unit  of 
length  for  dead,  and  m  for  live  load.  The  maximum  moment 
for  any  point  will  be  as  before. 

M—  —  -(j?-f  ra)  x  (£—#);     that  is,  will  be 

a 

given  by  a  parabola  whose  middle  ordinate  is  —  -  (p+m)  P. 

o 

For  the  shearing  force,  we  have 

Max.+S  =  Ip  (Z-2 
2 


Indicate  A  C  [Fig.  49,  PI.  14]  by  x^  for  which  max.—  S  =  0. 
then 

0  =p  I  (I—  2  Xi)—  m  a?!2, 


or 

m 


hence 


I          m          m2    m 

For  the  point  D  for  which  max.  +S  =  0,  B  D  =  a^.  The 
shearing  force  within  A  C  ia  positive,  within  B  D  negative, 
while  within  C  D  it  is  both  positive  and  negative. 

For        Z  =  5,      10,        20,         50,        75,        100,       150. 

5=0.12    0.19       0.31       0.64      1.05        1.55     3.12 
m 


:0.24    0.29      0.33       0.38       0.42       0.44     0.46Z 
CD  =-.0.52    0.42       0.34       0.24      0.16       0.12     0.08Z; 
that  is,  C  D  diminishes  with  increasing  span. 

§O.  Recapitulation.  —  For  girders  of  a  length  of  about  100 
feet  or  more,  then,  we  may  consider  the  live  load  as  distributed 
per  unit  of  length.  The  maximum  shearing  force  can  then  be 


96  SIMPLE   GIRDERS.  [CHAP.  VII. 

easily  found  according  to  the  preceding  Art.,  while  the  maxi- 
mum moments  will  be  given  by  the  ordinates  to  the  parabola 
for  full  live  and  dead  load  [Fig.  30,  Art.  44].  For  a  framed 
structure,  we  have  simply  to  multiply  the  shear  at  any  point 
by  the  secant  of  the  angle  which  the  brace  at  that  point 
makes  with  the  vertical,  in  order  to  find  the  strain  in  that 
brace.  The  moment,  divided  by  the  depth  of  truss  at  the  point 
in  question,  gives  the  strain  in  the  flanges.  For  opiate  girder, 
the  moment  being  found  as  above,  and  one  dimension  as  the 
depth  given,  we  can,  from  Art.  52,  so  proportion  the  other  di- 
mension as  that  the  strain  in  the  outer  fibre  shall  not  exceed 
the  amount  allowable  in  practice.  The  preceding  Art.  as  also 
Arts.  78  and  44  and  52  are  all  that  we  need  to  refer  to  for  all 
practical  cases  of  parallel  flange  girders  of  large  span. 

The  preceding  will  complete  our  discussion  of  the  simple 
girder.  We  have  only  to  remark  here  that  the  strains  due  to 
rolling  load  will,  in  general,  be  most  satisfactorily  found  by  the 
method  of  resolution  of  forces,  as  illustrated  in  Art.  12.  By 
this  method  we  first  find  the  reactions  at  the  supports  for  a  sin- 
gle apex  load,  either  graphically  or  by  a  simple  calculation 

,  and  then  follow  this  reaction  through  the 


girder,  arid  find  the  resulting  strains.  We  can  thus  find  and 
tabulate  the  strains  in  every  piece  due  to  a  weight  at  each  and 
every  apex.  The  maximum  strains  can,  then,  be  easily  taken 
from  the  table  thus  formed.  When  the  live  load  is  supposed 
thus  concentrated  at  each  apex,  it  is,  as  we  have  seen  in  Art.  12, 
unnecessary  to  follow  through  every  reaction.  The  reactions 
due  to  the  first  and  last  weights  are  sufficient  to  fill  out  the 
table.  For  solid-built  beams  or  plate  girders,  the  principles  of 
the  present  Chap.,  therefore,  come  more  especially  into  play. 
(See  also  remarks  at  close  of  Chap.  Y.) 

The  preceding  principles  will,  it  is  hoped,  be  found  sufficient 
to  enable  the  reader  to  find  the  maximum  moments  and  shear 
at  each  and  every  cross-section  of  a  beam  of  given  span  rest- 
ing simply  upon  two  supports,  and  acted  upon  by  any  given 
forces  or  system  of  forces  in  any  given  position.  The  reader 
will  do  well  to  take  examples  of  simple  trusses,  and  check  the 
results  obtained  by  the  method  given  in  Chap.  I.  by  the  above 
principles.  The  method  of  tabulation  of  single  apex  loads 


CHAP.  VII.]  SIMPLE    GIRDERS.  97 

upon  which  we  lay  so  much  stress  is  fully  given  by  Stoney 
["  Theory  of  Strains,"  vol.  i.],  and  the  examples  there  given  will 
be  found  of  service. 

Finally,  then,  the  strains  in  upper  and  lower  chords  are  great- 
est for  full  load  over  whole  span.  We  have,  therefore,  only  to 
erect  upon  the  given  span  a  parabola  whose  centre  ordinate  is 

— — -~- — ,  where p  is  the  load  per  unit  of  length  for  dead,  and 

8 

m  for  live  load  [Art.  44].  ,The  ordinates  to  this  parabola  at 
any  point  give  at  once  the  maximum  moment  at  that  point. 
The  depth  of  truss  at  this  point,  if  a  framed  structure,  or  the 
moment  of  inertia  of  the  cross-section  at  this  point,  if  it  is  a 
solid  beam  [Art.  52],  being  known,  the  strain  in  the  flanges  or 
outer  fibres  may  be  easily  determined.  The  strain  in  the  web  is 
given  by  the  maximum  shear.  For  dead  load  alone  this  is 
given  by  the  ordinates  to  a  straight  line  passing  through  the 

centre  of  span,  whose  extreme  ordinates  are^— -  [Art.  77].     The 

2i 

maximum  shear  due  to  live  load  alone- (m  I)  will  be  given  by 
the  ordinates  to  two  semi-parabolas,  convex  to  the  span,  having 

their  vertices  at  each  end,  and  the  extreme  ordinates  -   -   [Art. 

A 

78].  At  any  point,  the  greatest  of  the  two  ordinates  to  these  para- 
bolas is  to  be  taken.  For  live  and  dead  loads  together,  Art.  79 
may  also  be  useful.  The  shear  being  known,  the  strain  in  any 
diagonal  is  equal  to  the  shear  multiplied  by  the  secant  of  the 
angle  made  by  the  diagonal  with  the  vertical  [Art.  10  of  Ap- 
pendix] i.m  parallel  flanges.  For  flanges  not  parallel,  we  must 
find  the  resultant  shear  as  given  in  Art.  16  (4)  of  Appendix, 
or,  better  still,  the  flanges  once  known,  the  diagonals  can  be 
diagrammed  according  to  the  principles  of  Chap.  I. 

For  the  investigation  of  load  systems,  the  principles  of 
Arts.  70-75  will  be  found  sufficient,  and  the  application  of 
these  principles  we  have  already  sufficiently  illustrated  in  Arts. 
49-51. 

7 


$8  SUPPLEMENT   TO   CHAP.  VII.  [CHAP.   \. 


SUPPLEMENT  TO  CHAPTEK  VII. 


CHAPTER  l' 


METHODS  OF   CALCULATION. 

1. — In  Chapter  I.  of  the  text  we  hare  already  obtained  a  method  of  dia- 
gram which  will  be  found  both  simple  and  general,  and  by  which  we  can 
readily  determine  the  strains  for  any  given  loading  in  any  framed  struc- 
ture, no  matter  how  irregular  in  its  shape  or  dimensions,  provided  only  that 
all  the  outer  forces  are  Tcnown. 

In  Chap.  VII.  we  have  also  been  put  in  possession  of  another  method  of 
diagram,  by  which  we  may  for  any  structure  of  the  above  class,  framed  or 
not,  determine  the  moment  at  any  point,  and  can  then  properly  proportion 
the  cross-section. 

Thus  far,  indeed,  we  are  unable  to  apply  these  methods  to  the  continuous 
girder  or  braced  arch,  as  in  these  cases  there  are  not  only  upward  reactions 
but  also  end  moments,  and  in  the  latter  case  a  thrust  also,  which  must  first 
be  determined.  The  determination  of  these  requires  that  the  elasticity  of 
the  material  and  cross-section  of  the  structure  be  taken  into  account.  But 
with  these  exceptions,  and  they  are  of  rare  occurrence  in  practice,  we  can 
already  solve  any  case  which  may  present  itself. 

In  the  Appendix,  if  he  has  attended  to  our  numerous  references  to  it, 
the  reader  will  have  already  become  familiar  with  two  corresponding  meth- 
ods of  calculation,  viz.,  that  by  resolution  of  forces  and  that  by  moments. 

It  is,  however,  in  many  cases  desirable  to  know  not  only  the  strains  in 
every  piece  of  a  structure,  but  also  the  deflection  of  the  structure,  and  this 
also  requires  a  knowledge  of  the  theory  of  flexure  or  of  elasticity.  For  the 
sake  of  completeness,  therefore,  aiming  as  we  do  to  put  the  reader  in  pos- 
session of  methods  of  calculation  as  well  as  of  graphic  determination,  we 
shall  devote  a  few  pages  here  to  a  brief  notice  of  these  two  above-men- 
tioned methods  of  calculation,  and  then  pass  on  to  the  theory  of  elasticity 
itself.  This  latter  has  been  too  generally  considered  by  those  unacquainted 
with  the  methods  of  the  calculus  as  difficult  and  abstruse.  It  is  true  that 
the  calculus  must  be  called  into  requisition ;  but  so  simple  are  the  processes 
for  beams  of  single  span — and  it  is  with  these  only  we  have  at  present  to 
do — that  we  indulge  the  hope  that  by  going  back  to  first  principles  we  may 
enable  even  those  at  present  unacquainted  with  the  calculus  to  follow  our 


CHAP,  l']  SUPPLEMENT  TO  CHAP.  VII.  99 

I 

demonstrations  intelligently,  and  to  comprehend  perfectly  and  even  apply 
readily  the  method  for  themselves. 

We  cannot,  indeed,  make  the  reader  familiar  with  all  the  principles  of 
the  calculus,  but  all  these  principles  are  by  no  means  needed.  Its  funda- 
mental idea,  a  few  of  its  terms  and  applications,  are  all  that  he  need  be 
familiar  with  in  order  to  perform  the  simple  integrations  we  shall  encoun- 
cer,  as  readily  as  the  most  skilled  mathematician.  This  portion  of  the 
present  Supplement  may,  perhaps,  be  considered  by  many  as  unnecessary  and 
superfluous.  We  are,  indeed,  justified  in  assuming  such  knowledge.  But 
as  we  believe  our  plan  practicable,  we  cannot  resist  the  desire  of  making 
our  development  intelligible  to  all,  and  thus  rendering  our  treatment  of  the 
simple  girder  at  least  complete. 

The  practical  man  as  well  as  the  mathematician  may  thus  have  at  his 
disposal  the  powerful  aid  of  the  calculus,  so  far  at  least  as  his  purposes 
require  it,  and  be  able  to  deduce  for  himself  the  formulae  which  hitherto 
he  has  accepted  "  upon  faith."  It  may  also  not  be  improbable  that  here 
.  and  there  one  may  be  found  who,  pleased  with  the  simplicity  of  the  prin- 
ciples and  the  f ruitf  ulness  of  their  application,  may  be  led  to  further  prose- 
cute the  study  for  his  own  satisfaction. 

We  shall  first,  then,  notice  briefly  the  two  methods  of  calculation  above 
referred  to ;  then  devote  a  few  pages  to  the  development  of  those  prin- 
ciples and  rules  of  the  calculus  of  which  we  shall  make  use,  and  finally 
apply  these  principles  to  the  discussion  of  the  curve  of  deflection  of  loaded 
beams. 

2.  Hitter's  Method. — This  method  is  referred  to  in  Art,  14.  It 
rests  simply  upon  the  principle  of  the  lever,  or  the  law  of  statical  moments  ; 
requires  no  previous  knowledge,  and  converts  the  most  difficult  cases  of 
strain  determination  into  the  most  elementary  problems  of  mechanics. 
Ritter,  'in  his  "  Theorie  eiserner  Dach- und  Briicken-Constructionen,"  has 
applied  this  simple  principle  in  such  detail  and  fullness,  and  so  clearly  set 
forth  its  elegance  and  simplicity,  that  it  very  generally,  and  justly,  goes  by 
his  name. 

"  Its  results  are  clear  and  sharp  as  the  results  of  Geometry,  and  of  direct 
practical  application.  There  is  hardly  another  branch  of  engineering 
mechanics  which,  for  such  a  small  amount  of  previous  study,  offers  such 
satisfactory  results,  and  which  is  so  suited  to  engage  the  interest  of  the 
beginner." 

We  have  given  in  the  Appendix  to  Chap.  I.  (Arts.  6,  9,  10)  detailed  ex- 
amples of  its  application.  Throughout  this  work  similar  illustrations  of 
its  use  will  be  met  with,  so  that  it  is  only  necessary  here  to  state  more  fully 
than  in  the  text  its  general  principle. 

If  any  structure  holds  in  equilibrium  outer  forces,  it  does  so  by  virtue  of 
the  strains  or  inner  forces  which  these  outer  forces  produce.  Now  the 
outer  forces  being  always  given,  we  wish  to  find  the  interior  forces  or 
strains.  If,  then,  the  structure  is  framed,  and  we  conceive  it  cut  entirely 
through,  the  strains  in  the  pieces  thus  cut  must  hold  in  equilibrium  all  the 
outer  forces  acting  between  the  section  and  either  end.  Thus,  in  Fig.  6, 
PL  2,  a  section  cutting  D,  7  and  H  completely  severs  the  truss.  Then  the 


100  SUPPLEMENT   TO    CHAP.  VII.  [CHAP.  1.' 

strains  in  these  three  pieces  must  hold  in  equilibrium  the  reaction  at  A  and 
all  the  forces  between  A  and  the  section. 

Now  the  principle  of  statical  moments  is  simply  that,  when  any  number 
of  forces  in  a  plane  are  in  equilibrium,  the  algebraic  sum  of  their  moments 
with  respect  to  any  point  in  that  plane  must  be  zero. 

The  application  of  this  principle  is  simply  so  to  choose  this  point  of 
moments  as  to  get  rid  of  all  the  unknown  strains  in  the  pieces  cut,  except  one 
only ;  and  then  the  other  forces  being  known  in  intensity,  position,  and 
direction  of  action,  we  can  easily  find  this  one ;  since,  when  multiplied  by 
its  known  lever  arm,  it  must  be  equal  and  opposite  to  the  sum  of  the 
moments  of  the  known  forces. 

In  a  properly  constructed  frame  it  will,  in  general,  always  be  possible  to 
pass  a  section  cutting  only  three  pieces.  Then,  by  taking  as  a  centre  of 
moments  the  intersection  of  any  two,  we  can  easily  find  the  strain  in  the 
third. 

Even  if  any  number  of  pieces  are  thus  cut,  if  all  but  one  meet  at  a  com- 
mon point,  the  strain  in  this  one  can  be  determined. 

Thus,  in  Fig.  IV.,  PL  1  of  the  Appendix,  a  section  may  be  made  cutting 
2  3,  d  h,  h  c  and  c  Y.  But  all  these  pieces,  except  the  last,  meet  in  2,  and 
the  strain  in  this  last  piece  may,  therefore,  be  easily  determined. 

The  above  is  all  that  is  necessary  to  be  said  as  to  this  method.  The  ex- 
amples already  referred  to  will  make  all  points  of  application  and  detail 
plain  as  we  proceed.  We  see  no  reason  why  the  reader  who  has  mastered 
Chapter  I.  and  diligently  followed  out  the  examples  as  given  in  the  Appen- 
dix, should  not  now  be  able  to  both  calculate  and  diagram  the  strains  in 
any  framed  structure  all  of  whose  outer  forces  are  known. 

3.  Method  by  Re§olutioii  of  Forces. — We  have  also  yet  another 
method  of  calculation,  based  upon  the  principle  that,  if  any  number  of 
forces  in  a  plane  are  in  equilibrium,  the  sum  of  their  vertical  and  hori- 
zontal components  are  respectively  zero.  In  structures  all  the  forces  acting 
upon  which  are  vertical,  and  such  are  all  bridge  and  roof  trusses,  etc.,  of 
single  span,  we  have  only  to  regard  the  vertical  components. 

In  this  connection  we  have  to  call  attention  to  the  following  terms  and 
considerations.  The  shear  or  shearing  force  at  any  point  is  the  algebraic 
sum  of  all  the  outer  forces  acting  between  that  point  and  one  end.  These 
outer  forces  are  the  weights  and  reactions  at  the  ends.  At  any  apex  of  a 
framed  structure,  where  several  pieces  meet,  the  horizontal  components  of 
the  strains  in  these  pieces  must  balance,  or  the  structure  would  move ;  and 
for  the  same  reason,  the  algebraic  sum  of  the  vertical  components  must  be 
equal  and  opposite  to  the  shear.  The  shear  being  known,  if  the  strains  in 
all  the  pieces  but  one  are  also  known,  that  one  can  be  easily  found.  Thus 
the  algebraic  sum  of  all  the  vertical  components  of  the  strains  in  the  other 
pieces  being  found,  and  added  or  subtracted  from  the  shear,  as  the  case 
may  be,  the  resultant  shear,  multiplied  by  the  secant  of  the  angle  made  by 
the  piece  in  question  with  the  vertical,  gives  at  once  its  strain. 

This  method  is  also  fully  explained  in  the  Appendix,  Art.  16  (4),  and  a 
practical  rule  is  there  given  for  properly  adding  the  vertical  components 
and  determining  whether  the  result  is  to  be  added  to  or  subtracted  from 


J. 


CHAP.  I.]  SUPPLEMENT  TO  CHAP.  VII.  101 

the  shear.  This  rule  we  owe  to  Hiuriber*  We  have  thus  two  methods  of 
calculation,  which,  for  the  sake  of  convenience,  we  may  speak  of  as  Bitter's 
and  Humberts.  Corresponding  to  Humberts  method  we  have  also  a  graphic 
solution,  based  upon  the  same  principles  precisely.  This  we  have  set  forth 
in  Chapter  I.,  and  may  call  Prof.  Maxwell's  method.  In  Chapter  II.  and 
the  following  we  have  also  become  acquainted  with  the  graphic  solution 
corresponding  to  Hitter's  method,  or  the  method  of  moments,  which  we 
may  speak  of  as  Culmanrit*.  It  is  to  this  method,  based  upon  the  proper- 
ties of  the  equilibrium  polygon,  that  the  graphical  statics  properly  owes  its 
value  and  fruitfulness,  and  to  it  is  due  whatever  pretensions  it  can  claim 
as  a  system.  It  will  be  seen  hereafter  that  it  alone  can  furnish  a  general 
method  applicable  to  all  structures,  whether  framed  or  not ;  whether  all 
the  outer  forces  are  known  or  not.  By  the  same  general  method  we  are 
enabled  to  find  the  centre  of  gravity  and  moment  of  inertia  of  areas,  and 
to  solve  thus  a  great  variety  of  practical  problems —through  which,  how- 
ever different,  runs  one  universal  method,  one  simple  routine  of  construc- 
tion. 

*  Strains  in  Girders,  calculated  ~by  Formulas  and  Diagrams. 


102  SUPPLEMENT    TO    CHAP.  VII .  [CHAP.  II. 


CHAPTER  II. 

PRINCIPLES  OF   THE   CALCULUS   NEEDED  IN  OUR  DISCUSSION. 

4.  Differentiation  and  Integration.— We  need  but  a  very  few 
simple  ideas  and  conclusions  in  order  to  have  at  our  disposal  the  whole 
theory  of  flexure  for  beams  of  single  span.  Those  to  whom  these  ideas  are 
not  familiar  already  may  find  them  indeed  new,  but  will  not  find  them 
difficult  or  even  abstruse,  and  with  attention  to  the  following  will,  we 
venture  to  think,  make  a  valuable  acquisition. 


The  sign  /  is  called  the  "  sign  of  integration,1'  and  integration 


means 


simply  summation.  It  arises  merely  from  the  lengthening  of  the  original 
letter  S,  first  used  by  Leibnitz  for  the  purpose.  The  letter  d  is  called  the 
"sign  of  differentiation-;''  in  combination  with  a  letter,  as  d  x,  it  reads 
"  differential  of  #,"  and  signifies  simply  the  incitement  which  has  been 
given  to  the  variable  x.  So  much  for  terms. 

Now  suppose  we  have  the  equation 

y  =  5x\     .     .     .......     (1) 

in  which  x  and  y,  although  varying  in  value,  must  always  vary  in  such  a 
way  that  the  above  equation  holds  always  true.  This  being  the  case,  let 
us  give  to  y  an  increment  —  that  is,  supposing  it  to  have  some  definite  value 
for  which,  of  course,  x  is  also  definite  in  value,  increase  this  value  by  d  y. 

Then  x  will  be  increased  by  its  corresponding  amount  d  x,  and  as  the 
above  relation  must  always  hold  true,  we  have 

y  +  dy  =  5  (x  +  dxy    .     .     .     ...     .    (2) 

or  y  +dy  =  5  (x*+Zxdx  +  d  a;2). 

Inserting  in  this  the  value  of  y  from  (1),  we  have 

dy  =  5  (2xdx  +  dx*),     ......  (3) 

which  is  the  value  of  the  increment  of  y  or  d  y,  in  terms  of  x  and  the  in- 
crement of  x  or  d  x.  That  is,  the  increments  are  not  connected  by  the  same 
law  as  the  variables.  The  variable  y  is  always  5  times  the  square  of  the 
variable  x,  but  the  increment  of  y  is  greater  than  5  times  the  square  of  the 
increment  of  x  by  an  amount  indicated  by  5  x  2  x  d  x.  From  (3)  we 
have 


(4) 


which  gives  the  value  of  the  ratio  of  the  two  increments.  Now,  if  we 
assume  a  certain  value  for  x,  we  find  easily  from  (1)  the  corresponding 
value  of  y.  If  we  increase  this  value  of  x  by  a  certain  assumed  increment, 
d  x,  we  find  easily  from  (3)  the  corresponding  increment  of  y,  or  d  y.  Then 
(4)  would  give  us  the  ratio  of  these  two  increments. 


CHAP.  II.]  SUPPLEMENT  TO  CHAP.  VII.  103 

Now  we  see  at  once  from  (4)  that  the  smaller  we  consider  d  x  to  be,  the 
nearer  this  ratio  approaches  the  limiting  value  5  x  2  x.  We  may  suppose 
d  x  as  small  as  we  please,  and  then  this  ratio  will  differ  as  little  as  we 
please  from  5  x  2  a.  This  value,  5  x  2  x,  forms,  then,  the  limit  towards  which 

the  value  of  the  ratio  ~  approacfies  as  d  x  diminishes,  but  which  limit  evi- 

Ou  X 

dently  it  can  never  actually  reach  or  exactly  equal.  Because,  in  order  that 
this  should  bo  the  case,  d  x  must  be  zero.  But  if  d  x  is  zero,  that  is,  if  #  is 
not  increased,  y  also  is  not  increased;  d  y  is,  therefore,  zero,  and  there  is  no 
ratio  at  all. 

Now,  just  here  comes  in  what  we  may  regard  as  the  central  principle  of 
the  calculus. 

If  two  varying  quantities  are  always  equal  and  always  approaching  certain 
limits,  then  those  limits  must  themselves  he  equal. 

The  principle  is  too  obvious  to  need  demonstration.  "Two  quantities 
always  equal  present  but  one  value,  and  it  seems  useless  to  demonstrate 
that  one  variable  value  cannot  tend  at  the  same  time  towards  two  constant 
quantities  different  from  one  another.  Let  us  suppose,  indeed,  that  two 
variables  always  equal  have  different  limits,  A  and  B ;  A  being,  for  ex- 
ample, the  greatest,  and  surpassing  B  by  a  determinate  quantity  A. 

The  first  variable  having  A  for  a  limit  will  end  by  remaining  constantly 
comprised  between  two  values,  one  greater,  the  other  less  than  A,  and  hav- 
ing as  little  difference  from  A  as  you  please;  let  us  suppose  this  difference, 

for  instance,  less  than  —  A  .     Likewise  the  second  variable  will  end  by  re- 

6  f 

maining  at  a  distance  from  B  less  than  —  A.     Now  it  is  evident  that,  then, 

the  two  values  could  no  longer  be  equal,  which  they  ought  to  be  according 
to  the  data  of  the  question.  These  data  are  then  incompatible  with  the 
existence  of  any  difference  whatever  between  the  limits  of  the  variables. 
Then  these  limits  are  equal."  * 

Now  let  us  apply  this  principle  to  equation  (4).  In  this  equation  -y—  is 
a  variable  always  equal  to  5  (2  x  +  d  x).  But  5  (2  x+d  a>),  as  we  diminish 
d  x,  approaches  constantly  the  limit  5  x  2  x;  and  as  -= —  is  always  equal  to 

5  (2  x  +  d  x),  it  also  constantly  approaches  the  same  limit.     These  limits, 

dy 

then,  are  equal,  and  the  limit  of  -=—  =  5  x  2  x. 

d  x 

Now,  if  we  conceive,  and  such  a  conception  is  certainly  possible,  d  x  to 
lie  the  difference  between  x  and  its  consecutive  or  very  next  value,  such  that 
between  these  two  values  there  is  no  intermediate  value  of  d  x  ;  then  d  y 
will  be  the  difference  between  two  consecutive  values  of  y ;  and  regarding, 

then,  d  x  and  d  y  in  this  light,  -=—  will  be  the  limit  of  the  ratio  of  the  in- 

CL  X 

*  The  Philosophy  of  Mathematics.    Bledsoe. 


104  SUPPLEMENT    TO    CHAP.  VII.  [CHAP.  II. 

crements,  since  the  increments  are  then  limiting  increments,  and  can  be  no 
smaller  without  disappearing. 
We  have  thus 

^  =  5x2,, 
d  x 

which  is  an  exact  relation  between  the  increments  upon  this  supposition. 
From  this  we  have  d  y  =  5x2  x  d  x. 

If  now  we  sum  up  all  the  increments  d  y,  then  by  virtue  of  the  supposi- 
tion we  have  made,  /  d  y  must  equal  y.     We  thus  suppose  y  to  flow,  as  it 

were,  unbrokenly  along  by  the  consecutive  increments  d  y,  just  as  the  side 
of  a  triangle  moving  always  parallel  to  itself,  and  limited  always  by  the 
sides,  describes  the  area  of  that  triangle,  while  the  change  d  y  of  its  length 
is  the  difference  between  two  immediately  contiguous  positions.  Upon 

this  supposition,  we  repeat,  -= —   is  the  limit  of  the  ratio  of  the  increments, 
a  os 

which  limit  is,  as  we  see  from  (4),  equal  exactly  to  5  x  2  x.  We  do  not  re- 
ject  or  throw  away  d  x  from  the  right  of  that  equation  "  because  of  its 
small  size  with  reference  to  2  a,'1  but  simply  pass  to  the  limit,  and  then,  ac- 
cording to  our  fundamental  principle  above,  equate  those  limits  them- 
selves. But  if  /  d  y  =  y,  then  the  integral  of  5x2xdx,  or  /  5x2  xdx=y 

=5  &2.  By  "  differentiating,"  as  we  say,  equation  (1)  we  get  (5),  and  by 
"  integrating  "  (5)  we  obtain  (1). 

Hence  we  see  the  appropriateness  of  the  term  "fluent"  given  by  New- 
ton to  the  quantity  d  y  or  2  x  d  x.     So  also  we  see  the  appropriateness  of 

the  term  "ultimate  ratio  "  *  for  — -  itself. 

d  x 


*  Liebnitz  undoubtedly  discovered  the  calculus  independently  of  Newton, 
but  he  considered  d  x  as  a  quanity  so  "  infinitely"  small  that  in  comparison 
with  a  finite  quantity  it  could  be  disregarded  "  as  a  grain  of  sand  in  compari- 
son with  the  sea."  We  see,  indeed,  from  eq.  (4)  that  if  d  x  upon  one  side  be 

zero,  we  get  the  same  value  for  - —  as  before.     But  if  d  x  is  zero  on  one  side, 

it  should  be  zero  on  the  other  side  also.  No  matter  how  small  we  suppose  d  x 
to  be,  we  have  no  right  to  get  rid  of  it  by  disregarding  it.  That  Liebnitz  rec- 
ognized this  cannot  be  doubted,  and  he  was  therefore  inclined  to  consider  his 
method  as  approximate  only.  But  to  his  surprise  he  found  his  results  exact, 
differing  from  the  true  by  not  even  so  much  as  a  "  grain  of  sand."  There  was 
to  him  ever  in  his  method  this  mystery,  nor  could  he  conceive  what  these 
quantities  could  be  which.,  though  disregarded,  gave  true  results.  Bishop 
Berkeley  challenged  the  logic  of  the  method,  and  adduced  it  as  ac  evidence  of 
k' how  error  may  bring  forth  truth,  though  it  cannot  bring  forth  science." 
Strange  to  say,  even  the  disciples  of  Newton  were  unable  to  answer  Berkeley 
without  taking  refuge  in  the  undoubted  truth  of  their  results.  And  yet  New- 
ton in  his  Principia  lays  it  down  as  the  corner-stone  of  his  method,  that 
44  quantities  which  during  any  finite  time  constantly  approach  each  other,  and 


CHAP.  II.]  SUPPLEMENT  TO  CHAP.  VII.  105 

The  whole  of  the  calculus  is  but  the  deduction  of  rules  for  finding  from 
given  equations  as  (1)  their  "  differential  equations  "  as  (5),  or  inversely 
of  finding  from  the  differential  equation  by  "integration,"  or  summation, 
the  equation  between  the  variables  themselves. 

Such  of  these  rules  as  we  need  for  our  purpose  we  can  now  deduce. 

5.   Differentiation  and  integration  of  powers'of  a  single 

variable. — We  have  already  seen  that  the  /  d  y  =  y  and  /  2  xd  x  =  a?, 

hence  d  (x2)  =  2  x  d  x. 

If  we  should  take  y  =  x?>,  we  should  have,  in  like  manner,  as  before, 

y  +  d  y  —  (x  +  d  x)3  =  x*  +  3  a8  d  x  +  3  x  d  xz  4-  d  z3, 
or  dy=  3x*dx  +  3xdx2  +  dx*, 

or  ~  =  3z2  +  3xdx  +  dx'2, 

dx 

and  passing  to  the  limits,  as  before, 

- —  =  3  x2,  or  d  y=3  x2  d  x.    Hence  the  differential  of  x3  or  d  (a?3) =3  x2  dx, 

it   3} 

and  reversely,  the  integral  of  3  xz  d  x  or  /  3  xz  d  x  =  x3.  In  similar  man- 
ner, we  might  find 

d  (x5)  =  5  x*  d  x  and  A  x*dx  =  x5. 

Comparing  these  expressions,  we  may  easily  deduce  general  rules  which 
will  enable  us  at  once  upon  sight  to  "  differentiate,"  that  is,  find  the  rela- 
tion connecting  the  increments ;  and  "  integrate  "  or  sum  up  the  successive 
consecutive  values  of  the  variable;  for  any  expression  containing  the 
power  of  a  single  variable. 

These  rules  are  as  follows : 

To  differentiate: 

"  Diminish  the  exponent  of  the  power  of  the  variable  by  unity,  and  then 
multiply  by  the  primitive  exponent  and  ~by  the  increment  of  the  variable." 

Thus,  d(xz)  =  2xdx,  d  (x*)  =  3  x*  dx,  d  (a?7)  =  7*6  dx,  d(x$)  =  ^X2dx, 

d  (#")  —  n  a;0"1  d  x,  etc. 

To  integrate : 

"Multiply  the  variable  with  its  primitive  exponent  increased  "by  unity,  by 
the  constant  factor,  if  there  is  any,  and  divide  the  result  by  the  new  exponent." 

before  the  end  of  that  time  approach  nearer  than  any  given  difference,  an  eguaV 
There  can  be  little  doubt  that  Newton  saw  clearly  that  although  the  quantities 
might  never  be  able  to  actually  reach  their  limits,  yet  that  those  limits  them- 
selves were  equal,  and  hence  the  increment  could  be  left  out  in  the  equation, 
but  not  because  by  any  means  it  was  of  insignificant  size.  His  terms  ' '  ultimate 
ratio"  and  '•''fluent'1'1  are  alone  sufficient  to  indicate  that  he  understood  the 
true  logic  of  the  method  he  discovered ;  while  Liebnitz  seems  to  have  stood 
gazing  with  wonder  at  the  workings  of  the  machine  he  had  found,  but  whose 
mechanism  he  did  not  understand.  [See  Philosophy  of  Mathematics.  Bledsoe. 
Lippincott  &  Co.,  1868.] 


1U()  SUPPLEMENT   TO    CHAP.   VII.  [CHAP.  TI. 

Thus  A  xdx=2-j-=x*       f$afdx  =  -|"  =  *'       /V  dx  =  X~ 

/i            as*      2    f       /•                      n  x° 
x  dx=-  =  5-s        /  nx*-idx  = =  a»,  etc. 
i    a     /              » 

It  is  of  this  latter  rule  that  we  shall  make  especial  use  in  what  follows, 
6.  Other  Principles — Integration  between  limits,  etc. — 

We  may  observe  from  (1)  and  (4)  that  a  constant  factor  may  be  put  out- 
side the  sign  of  integration.     Thus    /  5x2  x  d  x  =  5    I  2  x  d  x  =  5  x\ 

It  is  also  evident  without  demonstration  that  the  integral  of  the  sum  of 
any  number  of  differential  expressions  is  equal  to  the  sum  of  the  several 
integrals. 

Thus  /  he  d  a  +  2s  d  z  +  y*dy  +  x* 

is  the  same  as  /  x  d  x  +    I  z3  d  z  +  I  y*  d  y,  etc. 

If  in  (1)  we  had 

y  =  5  x*+a, 

where  a  is  a  constant,  we  should  have 

y  +  d  y  =  5  (x  +  d  x)*  +  a  =  5  (x*  +  2  x  d  x+d  a?)  +  a, 

or  d  y  =  5  (2  x  d  x  +  d  afy  or  ^|  =  5  (2  x  +  d  x)  ; 

whence 

— -^  =  5x2  a?,  ordy  =  5x2xdx,  or 
d  x 

just  the  same  as  before. 

The  integral  of  this  will  then  be  y  =  5  x1  as  before,  whereas  it  should  be 
y  =  5  x*  +  a. 

If  two  differential  equations,  then,  are  equal,  it  does  not  necessarily  follow 
that  the  quantities  from  which  they  were  derived  are  equal. 

We  should,  then,  never  forget  when  we  integrate  to  annex  a  constant.  The 
value  of  this  constant  will  in  any  given  case  be  determined  by  the  limits 
between  which  the  integration  is  to  be  performed. 

We  indicate  these  limits  by  placing  them  above  and  below  the  integral 
sign.  Thus  the  integral  ofx*dx  between  the  limits  of  x  =  +  h  and  x= — h  is 

r+h  r       x* 

I  x3  dx.  If  we  integrate  a;2  d  x,  we  have,  then,  /  a2  d  x  =  —  +  C, 
J-h  J 

where  O  is  a  constant  whose  value  must  be  determined  by  the  conditions 
of  the  special  case  considered.     If  we  introduce  the  value  of  x  =  h  for  one 

limit,  we  have  —  +  O.     For  x  =  2  h  for  another  limit,  we  have  -—  +  O. 
We  have,  then,  two  equations,  viz. : 


when  x  —  h, 


c 

Jr~h  x  ~  3~ 


CHAP.  H.]  SUPPLEMENT  TO  CHAP.  VII.  107 


rx  =  2h 

JZ.J* 

n  the  other,  w 

rh    7 

x  =  h,    I  x'dx=- 
*s  h 


and  when  x  =  2h  I   x*dx=  — -  +  C ; 

and  by  subtracting  one  from  the  other,  we  have  for  the  integral  between 

r2h    7 

the  limits  x  =  2  h  and  x  —  li,    I   x'2  dx  —  — 7*s,  and  O  thus  disappears. 


We  have,  then,  only  to  substitute  in  succession  the  values  of  the  variable 
which  indicate  the  limits,  and  subtract  the  results. 
If  also  there  is  but  one  limit,  we  could  determine  O  if  there  were  also  a 

condition,  such,  for  instance,  as  that  /  x1  d  x  should  equal  h  when  x  =  2  h. 

The  ratio  — ^  is  called  the  "first  differential  coefficient ; "  if  it  were  to 
d  x 

be  differentiated  again,  the  next  ratio,  viz.,  that  of  the  differential  of  the 

differential  of  y  to  differential  of  or2,  or  — -,  is  the  "  second  differential  co- 

d  x 

efficient,"  and  so  on. 

dy 
Thus,  y  =  x5 ;  dy  =  d  (x5)  =  5  x*d x,  or.-=—  =  5 x*  ;  differentiating  again, 

— -i  =  20  x3  d  x,  or  — -  —  20  x3,  and  so  on  to  third  differential  coefficient,  etc. 
dx  dx* 

7.  Example. — As  an  example  of  the  application  of  our  principles,  let 
it  be  required  to  determine  the  area  of  a  triangle.  Let  the  base  be  b  and 
the  height  h.  Take  the  base  as  an  axis,  and  at  a  distance  of  x  above  the 
base  draw  a  line  parallel  to  &,  and  at  a  very  small  distance  d  x  above  this 
line  draw  another,  thus  cutting  out  a  very  small  strip.  (Let  the  reader  draw 
the  Fig.)  Now  for  the  base  y  of  this  strip  we  have  the  proportion  h  —  x  :  y 

7)  T  ~h  'JC  (L  i/1 

::  h  :  5,  or  y  =  b ,  hence  the  area  of  the  strip  is  b  dx — .      But 

n  n 

the  area  of  this  rectangular  slip  is  not  equal  to  the  area  of  that  portion  of 
it  comprised  within  the  triangle.  It  projects  over  at  each  end,  and  is, 
therefore,  somewhat  greater.  Thus  for  the  small  trapezoid  actually  within 
the  triangle  we  have  for  the  upper  side  y',  h—(x+d  x)  :  y'::h  :  b,  or  y'  = 

b— -  (x  +  d  x).    Hence  y—y'  — ,  and  the  area  of  the  projecting  portion 

h,  h 

of  the  rectangle,  that  is,  its  excess  over  the  trapezoid,  is  then  (y—y')  d  x,  or 

bdx"1     __      ,        ,,        bxdx     b  d  x*      ,          da      ,      bx    bdx 

Therefore,  bdx  — — —  =  d  «,  or  - —  =  b  —  — — ,  where 


h  h  Ti  d  x  h        h 

d  a  is  the  area  of  the  small  trapezoid  itself.  Now  these  latter  two  quanti- 
ties are  always  equal  for  any  value  of  d  x.  But  as  d  x  decreases,  one  side 

of  the  equation  approaches  the  limit  &— — — ,  and  — — ,  therefore,  approaches 

h  d  x 

this  same  limit.  The  rectangle  itself  is,  then,  the  limit  of  the  ratio  of  the 
area  of  the  small  trapezoid  to  its  height,  and  we  can  then  equate  the  limits 
themselves,  remembering  that  in  this  case  d  a  is  the  area  passed  over  by  the 


108  SUPPLEMENT    TO    CHAP.  VII.  [CHAP.  II. 

side  y  in  passing  from  one  position  to  the  consecutive  or  very  next.    We 
have,  then,  da  =  bdx — ,  and  if  we  integrate  this  expression,  that  is, 

sum  up  all  the  d  a's,  we  have  the  area  of  the  triangle.     Therefore, 

x  d  x  5  x1 


=ji    -IT-  )m.-ji+* 

where  O  is  the  constant  of  integration,  which  we  must  never  forget  to  annex. 
Now,  in  the  present  case  we  wish  to  sum  up  all  the  areas  d  a,  or  "  integrate," 
between  the  limits  x  —  o  and  x  =  Ji.  But  for  x  —  o,  A  must  be  zero,  and 
hence  we  have  O  =  o  for  the  condition  that  x  starts  from  the  base.  If  in 
addition  to  this  condition  we  make  x  =  h,  we  have  the  sum  of  all  the  areas 
between  x  =  o  and  x  =  h. 

,  ,      &  h      &  h 
A  =  &  h .  =  — ,  as  should  be. 

The  above  reasoning  is  somewhat  prolix. 

If  we  thoroughly  appreciate  that  d  x  is  the  difference  between  two  con- 
secutive values  of  x,  we  see  at  once  that  we  obtain  the  limiting  value  of  the 
rectangle  directly  by  multiplying  its  base  by  d  x.  The  sum  of  all  these 
must  be  the  area.  This  conception  ofdx  enables  us  to  curtail  much  of  our 
reasoning. 

Let  us  take  the  same  problem  again,  but  this  time  take  the  axis  through 
the  centre  of  gravity  of  the  triangle ;  that  is,  at  $h  above  the  base.  Then 
for  the  base  y  at  any  distance  x  above  this  axis,  we  have 

-A-av     "h'l  or     =  -l>-*J!L 
3  3         h 

Multiply  this  by  d  x  upon  the  above  conception  of  d  x,  and  we  have  at 
once  not  for  the  rectangle  upon  y,  but  for  its  limiting  value,  that  is,  for  the 
area  of  that  portion  of  the  rectangle  included  within  the  triangle, 
,  -.         2,        b  x  d  x 

~y        =3  h     ' 

Integrating  this,  then,  we  have 

•2  ,  &  x  d  x      2  ,        bx* 


where  O  is  a  constant  to  be  determined  by  the  limits  as  before.     For  one 

limit,  x  =  —  7i,  and  hence  we  have 
3 

A'=-l 

2 

For  the  other  limit,  x  =  +  —h,  and  hence  we  have 
3 


If  we  subtract  the  first  from  the  second,  O  disappears,  and  we  have  A  = 

9  1 

A"—  A'  =  —  &  h  =  —  &  A,  as  before. 
18  2 

o 

We  might  also  have  integrated  first  between  the  limits  x  =  0  and  x  —  —h. 

3 


CHAP.  H.]  SUPPLEMENT    TO    CHAP.  VII.  101) 

4 

For  x  =  0,  C  =  0,  and  the  area  above  the  axis  is  then  —  b  Ji.     For  x  =  0  and 

lo 

x  = h,  we  have  for  the  area  below  the  axis  —  —  b  h.     This  area  has  a  dif- 

3  18 

ferent  sign  because  below.  If  we  give  it  the  same  sign  as  the  other,  and 
then  add  it,  we  have  the  total  area.  If  it  also  had  been  above,  the  total 
area  would  have  been  the  difference.  Generally,  then,  we  subtract  accord- 
ing to  our  rule. 

8.  Significance  of  the  first  differential  coefficient.— Any 
equation  between  two  variables  of  the  first  degree  is  the  equation  of  a 
straight  line.  If  of  the  second  degree,  it  represents  one  of  the  conic  sec- 
tions, an  ellipse,  circle,  parabola,  or  hyperbola,  Of  a  higher  degree,  a 
aurve  generally.  If,  then,  we  take  the  axis  of  x  horizontal  and  y  vertical, 
and  if  d  y  and  d  x  are  the  consecutive  increments  of  y  and  x,  that  is,  the  dif- 
ference between  any  value  and  the  very  next,  the  ratio  ~  is  evidently  the 

d  x 

tangent  of  the  angle  which  a  tangent  to  the  curve  at  any  point  makes  with  the 
horizontal. 

If,  then,  we  make  —  =  0,  and  find  the  value  of  the  variabb  x  corrc- 
d  x 

spending  to  this  condition,  we  find  evidently  the  value  of  x  for  which  the 
tangent  to  the  curve  is  "horizontal.  If  now  the  curve  is  concave  towards  the 
axis,  this  value  of  x,  substituted  in  the  original  equation,  will  give  the  maxi- 
mum or  greatest  value  of  the  ordinate  y ;  because  for  the  point  just  one 
side  of  this  the  tangent  slopes  one  way,  and  for  the  point  just  the  other 
side  it  slopes  the  other.  The  point  where  the  tangent  is  horizontal  must 
then  be  the  highest. 

If  the  curve  is,  on  the  other  hand,  convex  to  the  axis,  the  value  of  x,  which 

makes  — -  =  0,  substituted  in  the  original  equation,  will  give  y  a  minimum 
d  x 

value  for  similar  reasons.  By  setting  the  first  differential  coefficient,  then, 
equal  to  zero,  we  may  find  that  value  of  x  which  corresponds  to  the  maxi- 
mum or  minimum  value  of  the  ordinate,  as  the  case  may  be.  In  the  case 
of  the  deflection  of  simple  beams  upon  two  supports,  the  curve  is  always 
concave  to  the  axis,  and  hence  we  obtain  by  this  process  always  the  maxi- 
mum deflection. 

The  above  comprises  all  the  principles  of  which  we  shall  make  use  in  the 
discussion  of  the  theory  of  flexure.  With  a  little  study,  we  believe  that 
any  one  familiar  with  analytical  operations,  even  although  he  may  never 
have  studied  the  differential  or  integral  calculus,  can  follow  us  intelligently 
in  what  follows.  Whatever  points  may  still  be  a  little  obscure  will  clear 
up  as  he  sees  more  plainly  than  now  their  application. 


110  SUPPLEMENT   TO    CHAP.  VH.  [CHAP.  III. 


CHAPTER    III. 


THEORY   OF    FLEXURE. 

9.  Coefficient  of  EIa§ticity.  —  Let  us  now  take  up  the  theory  of 
flexure,  and  see  if  it  is  not  possible  so  to  present  the  subject  that,  in  the 
light  of  the  preceding  principles,  we  may  be  able  to  solve  all  such  prob- 
lems as  present  themselves. 

If  a  weight  P  acts  upon  a  piece  of  area  of  cross-section  A,  and  elongates 
or  compresses  it  by  a  small  amount  I,  we  know  from  experiment  that, 
within  certain  limits,  twice,  three  times,  or  four  times  that  weight  will 
produce  a  displacement  of  21,  31,  4  I,  etc.  These  limits  are  the  limits  of 
elasticity.  Within  them  practically,  then,  the  displacement  is  directly  as 
the  force.  If  we  assume  this  law  as  strictly  true  for  all  values  of  the  dis- 
placement, and  if  we  denote  the  original  length  by  L,  then,  since  the 

force  per  unit  of  area  is  -r,  and  since  this  unit  force  causes  a  displacement 
I,  in  order  to  cause  a  displacement  L  equal  to  the  original  length,  this 

unit  force  must  be  T  time*  as  great,  or  equal  to  —  T.     This  force  we  call 
I  At 

the  modulus  or  coefficient  of  elasticity.     It  is  always  denoted  by  E.     Hence 


The  coefficient  of  elasticity,  then,  is  the  unit  force  which  would  elongate  a 
perfectly  elastic  body  BY  ITS  OWN  LENGTH.  It  is  a  theoretical  force  then  ; 
but  as  the  law  upon  which  its  value  is  based  is  true  practically  within  cer- 
tain limits,  by  experiments  made  within  those  limits,  knowing  P,  A,  and 
L,  and  measuring  I,  we  can  find  what  the  force  would  have  to  be  if  the  law 
were  always  true.  Such  experiments  have  been  made,  and  the  values  of  E 
for  different  materials  are  to  be  found  in  any  text-book  upon  the  strength 
of  materials. 
From  (6)  we  have  for  the  unit  force  of  displacement 


These  expressions  will  be  found  useful  as  enabling  us  to  replace  often 
expressions  containing  an  unknown  displacement  by  a  definite  or  experi- 
mentally known  value. 

1O.  Moment  of  Inertia.  —  This  is  also  a  convenient  abbreviation, 
and  enables  us  to  replace  unknown  expressions  by  a,  in  any  given  case, 
perfectly  determinate  value. 

The  moment  of  inertia,  with  respect  to  any  axis,  is  the  algebraic  sum  of  the 


CHAP.  III.]  SUPPLEMENT   TO    CHAP.  VII.  Ill 

products  obtained  ~by  multiplying  the  mass  of  every  element  of  a  given  cross- 
section  T)y  the  square  of  its  distance  from  that  axis. 

If  a  parallelogram  stand  on  end,  and  then  its  support  be  suddenly  pulled 
away  from  under  it,  it  will  fall  over  backwards.  But  to  knock  it  over 
thus  requires  force.  The  force  which,  in  this  case  overturns  it  is  that  of 
inertia.  At  every  point  of  the  surface  there  is,  then,  a  force  acting,  depend- 
ing upon  the  mass  of  this  point.  But  not  alone  upon  the  mass.  A  force 
at  the  top  acts  evidently  with  more  effect  to  turn  the  body  over  than  one  at 
the  bottom,  which  merely  tends  to  make  it  slide.  The  moment  of  each  ele- 
ment of  the  area  is.  then,  a  measure  of  the  force  which  at  each  point  causes 
rotation,  and  the  sum  of  these  moments  is,  then,  the  measure  of  the  over- 
turning action  of  the  whole  force  of  inertia  upon  the  surface.  The  moment 
of  this  latter  force,  or  the  sum  of  the  moments  of  the  moments,  is,  then,  the 
moment  of  inertia  of  the  cross- section.  Each  element  of  the  surface  must 
then  be  multiplied  by  the  square  of  its  lever  arm,  and  the  sum  of  all  the 
results  thus  obtained  taken.  In  other  words,  the  moment  of  each  element 
is  itself  considered  as  a  force,  and  then  its  moment  again  taken.  The  sum 
is  denoted  by  I. '  For  any  given  dimensions  and  axis  it  is  a  perfectly  defi- 
nite quantity,  and  may  thus  often  replace  expressions  containing  unknown 
quantities. 

The  principles  of  the  calculus  just  developed  will  enable  us  to  deter- 
mine it  in  some  cases,  at  least,  very  readily.  Its  value  for  various  forms  of 
cross-section,  in  terms  of  the  given  dimensions,  is  given  in  every  text- book 
upon  the  strength  of  materials. 

Let  us  suppose  a  rectangular  cross-section  of  breadth  5  and  height  h,  and 
take  the  bottom  as  axis.  The  area  of  any  elementary  strip  is,  then,  b  d  x. 
If  its  distance  from  the  bottom  is  x,  we  have  for  its  moment  b  x  d  x,  and  for 
its  moment  of  inertia,  then,  &  #2  d  x.  Integrating  this  expression,  we  have 

_  &> 
-    3    +0. 

This  integral  is  to  be  taken  between  the  limits  x  =  0  and  x  —  h.    For  x  =  0, 

1)  x'  d  x  =  0,  and  hence  C  —  0.    For  x  =  h,  then,  we  have -.     If  the  axis 

3 

had  been  taken  through  the  centre  of  gravity,  we  should  have  the  above 

integral  between  the  limits  +  -and—  — .     For+— we  have h  O.       For 

2  2  2  24 

—  — , +  O.     Subtracting  one  from  the  other  (Art.  6),  we  have  

for  the  moment  of  inertia.  For  a  triangle  of  height  h  and  base  Z>,  we  have 
for  axis  through  centre  of  gravity,  from  Art.  7,  for  the  area  of  the  very 

small  strip  at  distance  a?,  -Idx  —  -  x  d  x.    Multiplying  this  by  x*,  we  have 

o  fl 

for  its  moment  of  inertia  -lxldx  —  -x*dx.       The  integral  of   this  is 
o  h 

2  T)  x4 

O  A 

For  x  =  —  h,  this  becomes  I 

o  243 


112  SUPPLEMENT   TO    CHAP.  VII.  [CHAP.  ITT. 

For  x  =——71,  we  have  —  — —  &  7i3  +  C. 

27  1 

Subtracting  one  from  the  other  (Art.  0),  we  have  -^  &  A3,  or  _—  &  A3  for 

the  moment  of  inertia.     The  moment  of  inertia  of  the  rectangle  I  =  — 

may  be  written  — =  — — -  x—  Ax    — ,  or  the  moinent  of  inertia  of  the  half 

parallelogram  is  equal  to  its  area,  into  the  distance  of  its  centre  of  gravity 
multiplied  by  f  ds.  of  its  height.  We  see  at  once  that  when  we  consider, 
then,  the  statical  moments  as  themselves  forces,  the  centre  of  action,  of  these 
moment  forces  does  not  coincide  with  the  centre  of  gravity  of  the  area.  This 
principle  we  have  already  noticed  in  Chap.  VI.,  Art.  60. 

I        A2        /I  \2  1 

We  can  also  put  --=_  =  (_    -  A )  .     This  value A  is  called  the 

A      12        \2V3    )  3y£ 

radius  of  gyration.  It  is  evidently  the  distance  from  the  axis  to  that  point 
at  which,  if  the  mass  were  concentrated  or  sum  of  all  the  forces  were  con- 
sidered as  acting,  their  moment  of  inertia  would  be  that  of  the  cross-sec- 
tion itself.  The  value  of  —  is,  in  general  then,  the  square  of  the  radius  of 

JA 

gyration.  We  have  already  shown  in  Chap.  VI.  how  to  find  it  graphically 
for  various  cross-sections. 

We  are  now  ready  to  take  up  the  case  of  a  deflected  beam,  and  to  find 
the  differential  equation  of  its  curve  of  deflection. 

11.  Change  of  SUiape  of  the  Axis. — In  the  Fig.  given  in  the 
Supplement  to  Chap.  XIV.,  we  have  represented  a  beam  deflected  from  its 
original  straight  line  by  outer  forces.  Let  the  two  sections  A  C,  B  D  be 
consecutive  sections,  parallel  before  flexure,  and  remaining  plane  after.  Let 
the  length  of  the  axis  m  a  be  s,  then  n  a  —  d  s,  and  let  d  <p  be  the  very 
small  angle  between  the  sections  after  flexure. 

If  the  deflection  is  small,  s  will  be  approximately  equal  to  x,  and  d  s  to 
d  x.  The  elongation  of  any  fibre  at  a  distance  «  from  the  centre  is.  then, 
v  d  (f).  The  unit  force  corresponding  to  this  elongation  is  from  (7)  T  = 

E  — -  v.     If  d  a  is  the  cross-section  of  any  fibre  as  d  c,  then  the  whole  force 
d  x 

of  extension  is 

Ev da d  $ 

dx 

The  moment  of  this  force  is,  then,  —  — = — - — -.  The  integral  of  this  be- 
tween the  limits  +  --  and will  give  the  entire  moment  of  rupture.  But 

this  is  equal  and  opposite  to  the  moment  M  of  ail  the  outer  forces ;  hence 


CHAP.  III.]  SUPPLEMENT   TO   CHAP.  VII.  113 

But,  as  we  have  just  seen,  this  integral  is  the  moment  of  inertia  I  of  the 
cross-section  with  reference  to  the   axis    through  the    centre.      Hence, 

M  =  — - — ® .     Since  rf>  is  a  very  small  angle,  it  may  be  taken  equal  to  its 
d  x 


tangent,  or  equal  to        ;   hence  -      and   M  =  B  I         . 

dx  d  x      d  or  d  x* 

But  v  d  cf) :  v  ::  d  x  :  r,  where  r  is  the  radius  of  curvature ; 

v  d  &      d  x  d  <b       1 

hence  • =  —  or  -r-^  =  — . 

v  r  dx      r 

Therefore,  M  =  Hli  =  EI^|=—  ;*  v  .    *    .     (8) 

r  dx*        v 

and  T  =  5-B  (9) 

r 

Equation  (8)  is  our  fundamental  equation. 

In  any  given  case  we  have  only  to  write  down  the  expression  M  for  the 

dz  y  ! 
moment  of  the  outer  forces  at  any  point,  and  equate  it  with  E  I  — -^. 

•     ••  Cu   <Xs      , 

Integrating  once  we  shall  then  have  for  I  constant,  of  course,  E  I  — -  and, 

d  x 

integrating  again,  E  I  y  in  terms  of  a?,  or  the  equation  of  the  deflection 
curve  itself.  Making  E  1~  =  0,  we  can  then  find  the  point  of  maximum 

deflection,  and  inserting  in  the  value  for  Ely  the  value  of  *  thus  found, 
can  find  the  maximum  deflection  itself.  The  discussion  of  any  case  reduces 
thus  to  a  simple  routine,  and  every  case  is  in  many  respects  but  a  repetition 
of  the  same  processes. 

12.  Beam  fixed  at  one  end  and  loaded  at  the  other— 
Constant  cro§s-section. — We  shall  always  consider  a  moment  positive 
when  it  causes  compression  in  the  lower  fibre ;  negative  when  it  causes  ten- 
sion in  that  fibre.  Distances  to  the  right  of  the  origin  are  always  positive, 
to  the  left  negative.  Hence  on  the  left  of  any  section  an  upward  force  is 
negative,  a  downward  force  positive ;  while  on  the  right  of  the  section  the 
upward  force  is  positive  and  the  downward  one  negative.  The  reader 
should  always  draw  the  Fig.  for  each  case  discussed,  and  in  the  beginning, 
at  least,  review  these  conventions  each  time. 

Now  let  a  beam  of  length  I  have  the  weight  P  at  the  free  end,  and  let  it 
be  fixed  horizontally  or  "  walled  in  "  at  the  right  end.  Then  the  moment 
at  any  point  distant  x  from  the  left  or  free  end  is  M  —  +  P  x. 

(a)  Change  of  shape. 

From  (8)  we  have  now 


114  SUPPLEMENT   TO   CHAP.  VII.  [CHAP.  III. 

Integrating  once  (Art.  5)  we  have 

BI^L^l 
dx          2 

where  C  is  the  constant  of  integration  to  be  determined  (Art.  6)  by  the 

given  conditions.      Now  by  the  condition  in  this  case,  when  x  =  I,  —  - 

a  x 

must  be  zero,  because  the  end  is  fixed,  and  the  tangent  there  must  therefore 

P  1? 
be  horizontal  (Art.  8).     Hence  O  =  —  -  ,  and 

E  2 


_ 
dx   ~      2          2' 

We  have  thus  introduced  the  condition  that  x  cannot  be  greater  than  I. 
Integrating  again  (Art.  5) 

P  a?      PVx 


Here  again  we  have  a  constant  to  be  determined,  and  here  again  we  have 
the  condition  that  for  x  —  l,y  must  be  zero,  since  at  the  fixed  end  there  can 

P  1? 

be  no  deflection.    Therefore,  O  =  — -    and 

o 

P  /  \       P 

E  I  y  ==  —  [2  lz  —  3  Z2  a;  +  a;8  i  = —  (21  +  x)  (I  —  a)2. 
6  y  j      6 

The  deflection  will  evidently  be  greatest  at  the  free  end,  and  here,  therefore, 
for  x  =  0,  we  have 

_       _  p*3  . 

If  the  cross-section  is  rectangular,  I  =  —  5  A3  (Art.  10),  and  the  maximum, 
deflection  A  =  — — — -• 


(5)  Breaking  weight. 

T  I 

We  have  also  from  equation  (8)  M  = ,  where  T  is  the  tensile  strain 

t> 

in  any  fibre  distant  «  from  the  centre.     For  v  —  — ,  T  is  the  tensile  strain  in 

2 

O     *p     T  Jj 

the  outer  fibre,  and  M  =  .     For  v  =  —  —  we  have  the  compressive 

7i  2 

2  C  T 

strain  in  the  outer  fibre  upon  the  other  side,  or  M  =  — - — .      Theoretically 

the  two  should  be  equal.     Practically  they  are  not.     In  fact,  if  we  put  for 

.Pa  = 
Tbh2 


2  T  I 

M  its  value,  we  have  P  x  =  — - — ,  or  for  a  rectangular  cross-section  P  x  = 

h 


i—  T  &  h2.    This  is  greatest  for  x  =  I,  hence  the  breaking  weight  P  =        •    . 

ft     TTJ      7 

From  this  we  have  T  =  -r-rr'     Now  experimenting  with  beams  of  various 
o  fi 


CHAP.  III.]  SUPPLEMENT   TO    CHAP.  VII.  115 

materials,  known  dimensions  and  given  weights,  we  may  find  experimen- 
tally T.  It  would  seem  that  this  value  thus  found  should  equal  either  the 
tenacity  or  crushing  strength  of  the  material,  but  the  results  of  experiment 
show  that  it  never  equals  either,  but  is  always  intermediate  between  T  and 
C.  Calling  this  intermediate  value  R,  we  have 


The  formula  is  based  upon  the  condition  of  perfect  elasticity,  while  R  is 
determined  by  experiments  made  at  the  breaking  point  when  the  condition 
of  perfect  elasticity  is  no  longer  fulfilled.  In  the  following  table  the  tabu- 
lated values  of  R  are  correct  for  solid  rectangular  beams,  and  sufficiently 
exact  for  those  which  do  not  depart  largely  from  that  form.  If  instead  of 

we  use  the  values  of  T  or  C,  whichever  is  the  smaller,  we  shall  always 
T)e  on  the  safe  side,  since  R  is  invariably  intermediate  between  these. 

In  general  we  shall  refer  to  the  equation 


when  we  have  occasion  to  find  the  breaking  strength.     But  it  must  be 
always  remembered  that  in  any  practical  example  we  should  replace  T  by 
R  for  rectangular  beams,  or  by  T  or  O,  whichever  is  the  smaller,  for  others. 
We  give  also  the  values  of  the  coefficient  of  elasticity  E.     (Wood's  Resist. 
of  Materials.) 

TOR  E 

Cast-iron  .......................  16.000        96,000        36,000        17,000,000 

Wrought-iron  ..................  58^200        30,000        33.000        25,000,000 

English  Oak  ....................  17,000          9,500        10,000          1,451,200 

Ash  ............................  17,000          9,000        10,000          1,645,000 

Pine  ...........................     7,800          5,400          9,000          1,700,000 

All  in  pounds  per  square  inch. 

2.  Beam  of  uniform  strength. 

Suppose  the  cross-section  or  I  is  not  constant,  but  varies  so  that  at  every 
point  the  strain  T  is  constant.     From  (11)  we  have 

2  T  I 

M  =  P  x  —  —  ,'  -  for  the  outer  fibre,  whence 
h 


T  =  —  -.     For  a  rectangular  cross-section  T  =   ,      '  .     Now  suppose  the 
21  o  h 


' 
2 

breadth  and  height  at  the  fixed  end  are  &i  and  hi.     Then  at  this  end  T  = 
n  ~p  7 


But  this  must  be  equal  to  T  at  any  other  point  ;  hence 

6Pa        QPl         IV        x 
i  A,2 


If  we  suppose  the  height  constant,  we  have  for  the  varying  breadth  at  any 

point  b  =  li  X.    That  is,  the  breadth  must  vary  as  the  ordinates  to  a  straight 
I 

line,  and  the  plan  of  the  beam  is  a  triangle  with  the  weight  P  at  the  apex. 


116  SUPPLEMENT   TO    CHAP.  VII.  [CHAP.  III. 

If  the  breadth  is  constant,  h  =  hi./  £.,  or  the  elevation  of  the  beam  is  a 
parabola  with  the  weight  at  apex.     If  the  cross-section  is  always  similar, 

that  is,  if  —  =  -,  we  have  5  =  -\—,  and  substituting  in  the  equation  above 
hi       h  hi 

h  =  hi  ?  /  £.  ,  which  is  a  paraboloid  of  revolution. 

(a)  Change  of  shape 
From  (8)  we  have 

dz  y  _  P  x  _       P  x 

where  &  and  h  are  variable.     If  we  suppose  the  height  h  constant  and 
always  equal  to  hi,  then,  as  we  have  seen,  &  =  &i  yj  hence  for  rectangular 

cross-section 

d2  y  _    12  PI 
d  x2      E  hi*  5i 

/I    nj 

Integrating,  since  for  x  =  I,  — —  =  0,  we  have 

(L  (K 


dy  _  12  P  Ix  _  12  F  Z2 
d  x       E  hi*  &i        E  hi3  &i* 
Integrating  again,  since  f or  x  =  Z,  y  =  0.  we  have 
_  6P?a2  _12PZ2a          6P 
E  hi3  &i         E  hiz  bi         E  h^ 

For  the  maximum  deflection  x  —  0,  and 

.  ^  ^^ 


The  above  value  of  y  can  be  written 


but  is  —  ,  the  deflection  of  a  beam  of  constant  cross-section  &i  Alt 

as  already  found.     Calling  this  deflection  A0,  we  have 


Q 

for  the  deflection  at  any  point,  or  A  =  —  A0  for  the  maximum  deflec- 

tion. 

In  a  similar  manner,  for  constant  breadth,  we  have 


CHAP.  HI-.]          SUPPLEMENT  TO  CHAP.  VII.  117 

For  similar  cross-sections,  we  have 

9        r         5  x       3  „  I       \5  ~  9  36       P  I3 


If  we  call  the  volume  of  the  beam  of  constant  cross-section  V,  then  in 

1  2 

the  first  case  the  volume  Vj  =  —  V  ;  in  the  second,  V2  =  —  V  ;  in  the  third, 

6  O 

V3=?-V;    or 

V  :  V2  :  V3  :  V:  =  30  :  20  :  18  :  15. 
The  maximum  deflections,  as  we  see  above,  are  as 

2  Ao,  -  Ao,  -  Ao,  or  as  20,  18,  and  15. 

o         /* 

That  is,  the  deflections  at  the  ends  for  a  beam  of  uniform  strength  in  the 
three  cases  are  as  the  volumes. 

1.3.  Beam  a§  before  fixed  at  one  end—  Uniform  load- 
Constant  cro§s-sectioii.—  If  p  is  the  load  per  unit  of  length,  we  have 
for  the  moment  at  any  point  distant  x  from  the  free  end, 

x        p  xz  p  xz       dz  y 

M  =  *»x  Sf  =  -a-.and  hence  —  —  =^r. 

;.:.,.  .  p  p 

This  moment  is  greatest  for  x  =  I,  and  hence  Max.  M  =  —  —  . 

For  the  breaking  weight,  then,  from  (11) 

pi2      2TI  4  T  I 

T  =  —  or  pl=-hT> 

or  twice  as  great  as  for  an  equal  weight  at  the  end. 
For  the  change  of  shape,  we  integrate  twice,  precisely  as  before,  the  ex- 


»          d2  y       p  xz 
pression    -=  —  j  —  and  obtain  thus 


*       24  B  I 
The  maximum  deflection,  then,  is 


_  4  p  x 


or 


A  =  8EI' 

or  only  fths  as  great  as  for  an  equal  load  at  the  end. 
2.  €on§tant  strength. 

We  have,  as  before,  from  (11)  M  =  *£  =  ^,  whence  T  =  ^f- 
for  rectangular  cross-section,  I  =  i  I  &  and  T  =  *-~.     If  h  A,  are  the 

breadth  and  heighth  of  the  fixed  end  section,  then,  since  T  must  be  always 
constant, 

3  px2  _  3p  I2        I  h*  __  x2  • 

5  A2  far*?  or&7^7=  Is"" 


For  lieight  constant,    5  =  &j  |  X-  \ 


118  SUPPLEMENT   TO   CHAP.  VH.  [CHAP.  HI. 

<y* 

For  breadth  constant,    h  =  hi  -=• . 

,  V 

For  similar  cross -sections,   h  =  fa 

The  first  is  in  plan  a  parabola ;  the  second,  in  elevation  a  triangle ;  the 
third,  a  paraboloid  of  revolution. 

For  the  change  of  shape,  -we  have,  by  proceeding  in  the  same  manner  as  in 
Art.  12,  A  =  2  Ao,  A  =  4  Ao,  and  A  =  3  A0  in  the  three  cases,  where  A0  is  the 
deflection  of  a  similar  beam  of  constant  cross-section  &i  hi. 

14.  Beam  supported  at  both  the  ends — Constant  cross- 
section — Concentrated  load. — Let  the  weight  P  be  distant  from  the 
left  end  by  a  distance  Zi  and  from  the  right  end  by  Za.  Let  the  distance  of 
any  point  from  the  left  end  be  x.  For  the  upward  reaction  at  the  left  end, 

—  P  —      k 

T 

The  moment,  then,  at  any  point  between  the  left  end  and  F,  for  x  less 
than  Zi,  isM  —  —  ""?""•  ^or  any  point  to  the  right  of  P,  or  #  greater 

than  llt  M'  =  . — +  P  i  x  —  Zi  I.     Instead  of  this,  however,  we  may 

take  the  reaction  at  the  other  end,  Va  =  P~;  and  then  for  x  greater  than  Zi, 


The  moment  is  evidently  greatest  at  the  point  of  application  of  the  load, 
or  for  x  =  li    Hence  the  maximum  moment  is  —  —  T—  -. 

(a)  Breaking  weight. 

2  T  I  P  I   I 

From  (11)    M  =  -   —  =  --  '—  ,  or,  for  the  breaking  weight,  P  = 

2  T  I  Z  1  T  5  A?  Z 

-,  7   7  «    For  rectangular  cross-section,   I  =  77:  5  h*  and   P  —     a  -,   7    - 
ti  ti  It  t  12  o  Li  L-i 

For  a  load  in  the  middle,  Zi  =  Z2  =  —  I  and  Max.  M  =  —  -r  P  I,  and  P  = 

&  4: 

r  4  times  as  great  as  for  a  beam  of  same  length  fixed  at  one  end 
and  free  at  the  other. 
(5)   Change  of  shape. 
We  have,  then,  from  (8),  for  x  less  than  Zi, 

&  y          P  Z2  x       ,  &  y'          P  |j  (7—  x)  £  .,       , 

d*  =  -HIT  and  5/      "irr2  for  z  greater  than  '•• 

Integrating,  we  have 

•        d+y_      PZ,**  dtf  _       PI,    r        *fl,0. 

-~  =   "  " 


For  x  =  lit  these  two  values  of  —  ^-are  equal,  and  hence,  since  Z,  =  I  —  llt  we 

a  a; 


CHAP.  III.]  SUPPLEMENT   TO    CHAP.   VII.  119 

P  I  2 

have  C'  =  C  H — -^-.     We  have  then  the  two  equations 
d  y          Pl*x2  ,  d  y'          P  I, 

^=-^TI+C  and  jf  •  -«i 

containing  both  the  same  constant  C. 
Integrating  these,  we  have 

P  Z2  y?  P  Zi    rZ  a;2     a*n      P  Z,2  0 


In  the  first  of  these,  f  or  x  —  0,  y  —  0 ;  hence  d  =  0. 

P  Zi3 
For  x  =  Zi,  y  =  y' ;  and  hence  O2  =  —  . 

D  £!  I 

For  x  —  Z,  y'  —  0 ;  and  hence,  finally,  C  =        * '  (2  Z  —  Zi] 
We  have,  therefore,  by  substitution  of  these  constants, 


P  Zi2  Z22 
For  x  —  Zi,  we  have  the  deflection  at  the  load  y  =  —      —7. 

O  E  I  6 

>7  rti 

Inserting  the  value  of  C  in  the  value  for  — —  above,  and  placing  the 

value  of  —  equal  to  0,  we  have  for  the  value  of  #,  which  makes  y  a  maxi- 
d  x 

mum,  x  =*  /— (2  Z— Zj)  Zi,  an  expression  holding  good  only  for  x  less  than 

Zj.     Inserting  this  in  the  value  for  y,  we  have  for  the  maximum  deflection 
itself 


If  the  load  is  in  the  middle,  we  have  for  the  curve  of  deflection 


P  Z3 
and  for  the  deflection  itself  A  = 


48  El 

The  greatest  deflection  is  not,  then,  at  the  weight,  except  when  the  load  is 
in  the  middle.  When  this  is  the  case,  the  deflection  is  only  ^-th  of  the 
deflection  for  the  same  length  of  beam  fixed  at  one  end  and  loaded  at  the 
other  free  end. 

15.  Beam  as  before  supported  at  the  end§  —  Uniform 
load.—  For  a  load  p  per  unit  of  length,  the  entire  load  is  p  Z.  The  reac- 

tions at  each  end  are  —  -,  and  the  moment  at  any  point  is 


is  evidently  greatest  at  the  centre,  and  hence 
Mar.  M  =  -££ 


120  SUPPLEMENT   TO   CHAP.  VH.  [CHAP.  IH. 

For  the  breaking  weight,  then,  from  (8) 


16  T  I 


or  4  times  as  much  as  for  a  beam  of  same  length  loaded  uniformly  and 
fixed  at  one  end. 
For  the  change  of  shape,  we  have 

dzy  _^px  (l—x} 

dx*~          2EI 
The  constants  of  integration  are  determined  by  the  conditions  that,  for 

X  =  W'dx=®'  ^  —  °'  y~®'->  and  x  =  l,y  =  0.    Integrating,  then,  twice 
under  these  conditions,  we  have 


This  is  greatest  at"  the  centre,  or  for  x  =  —  ;  hence  the  maximum  deflection  is 

2 

A  =  —  -  •—-,  or  only  i&ths  of  a  beam  of  the  same  length  fixed  at  one 

end  and  uniformly  loaded. 

16.  Beam  supported  at  one  end  and  fixed  at  the  other 
—  Constant  cross-section—  Concentrated  load.  —  Let  the  left 
end  be  fixed  horizontally  so  that  the  tangent  to  the  deflected  curve  at  that 

point  is  always  horizontal,  and  therefore  —  =  0. 

a  x 

Let  the  distance  of  the  weight  P  from  left  be  a,  and  the  distance  of  any 
point  x. 
Then,  for  x  less  than  a,  we  have 

M  =  —  V  (l-x)  +  P  (a-x)  • 
for  x  greater  than  a, 

M'  =  -V(Z-aO, 
where  V  is  the  reaction  at  the  free  end,  and  is  so  far  unknown. 

If  we  put  M  =  —  -£  and  M'  =       '{,  and  integrate  as  usual,  and  remem- 
d  x  d  x* 

ber  that  f  or  x  =  0,  ^  =  0,  and  for  x  =  a,  -^  '  =  ~,  we  have 
d  x  dx      dx* 


Integrating  again  and  determining  the  constants  by  the  conditions  that, 
for  x  =  0,   y  =  0,   and  for  x  =  a,   y  —  y\  we  have 

V  =  ~~  [V  (3  «-aO-P  (3  «-*)] 

y1  =  -^  [V  *«  (3  Z_a)-P  (3  x-a}  a*]. 


CHAP.  III.]  SUPPLEMENT   TO    CHAP.  VII.  121 

Now,  for  x  =  I,  y  =  0  ;  hence  V  =  P  —  L_  —  -  -.     If  the  load  is  in  the 


middle,  V  =  —  P. 

,    .  V,  or  the  reaction  at  the  free  end,  is  now  known,  and  substituting  it  in 
,  the  value  of  y'  above,  we  have  the  equation  of  the  deflection  curve  between 
the  weight  and  the  free  end. 

y'  —  — — --\ —  —  (3  x  —  a)  a2 

Substituting  it  also  in  the  value  of  — —  above,  and  placing  then  — ^-  equal 

dx  dx 

to  zero,  we  find  for  the  value  of  x,  which  makes  the  deflection  a  maximum, 

when  x  is  greater  than  #,  x  =  I — 1\/  — j . 

'    3  &  —  o> 

Substituting  this  value  of  x  in  the  value  of  y'  above,  we  have  for  the 
maximum  deflection  itself 


P  Z3         1 
When  the  weight  is  at  the  middle,  this  becomes  A  =  -  x  —  =  ,  or 

48  E  I      y/5 

only  —  —  =  ,  as  much  as  for  a  beam  of  same  length  fixed  at  end  and  with 
tC  y  5 

load  at  other  end,  and  only  —  -  as  much  as  for  same  beam  simply  sup- 

y5 
ported  at  ends. 

Breaking  weight. 

Having  now  V,  we  know  M  and  M'.  Rupture  will  occur  where  the 
moment  is  greatest,  that  is,  either  at  the  fixed  end  or  at  the  weight.  Now 
the  moment  at  P  is  —  V  (I—  a)  —  —  V  1  +  V  a.  The  moment  at  the  fixed 
end  is  —  V  l  +  P  a.  Now,  as  V  is  always  less  than  P,  we  see  at  once  that 
for  any  value  of  a  less  than  I,  the  moment  at  the  weight  is  greatest. 

We  have  for  the  moment  at  the  weight  from  (8) 

2(3Z~a)/7      ,      2TI         ... 
i—j  —  -  (I-  a)  =  —  —  ,  and  hence 


4  T  1  13 

for  the  breaking  weight  P  =    -     —  — 


ha*  (3  I— a)  (l-o)* 

64  T  I 

If  the  weight  is  in  the  middle,    P  —  —  -=-.-, 

o    fi  l 

or  fths  as  much  as  for  the  same  beam  supported  at  the  ends. 

IT.  Beam  as  before  fixed  at  one  end  and  supported  at 
the  other — Uniform  load. — In  this  case  the  moment  at  any  point  is 

M—  —  V  (l—®)-r-p  (l-x)*  =  E  I  --Jj.     Integrating  twice  and  determin- 
ing the  constants  by  the  conditions  that,  for  x  =  0,  — -  =  0  and  y  =  0,  we 

d  x 

easily  obtain 


122  SUPPLEMENT   TO    CHAP.  VH.  [CHAP.  III. 

(B  l-**-P  (6  Z2~ 


o 

For  x  =  1,  y  =  0,  and  hence  V  =  —  p  1. 

8 

Substituting  this  value  of  V 


•j  r  _   V'-iR 

Putting  this  last  equal  to  zero,  we  find   x  —  —  -  -  Z,     or    x  =  0.5785  Z, 

lo 

for  the  value  of  a*,  which  makes  the  deflection  a  maximum,  and  this  in- 
serted in  the  value  of  y  gives  for  this  maximum  deflection  itself, 


164  El  El 


For  the  "breaking  weight,  we  have,  since  the  greatest  moment  is  at  the  fixed 

1      72    __      1      72       2  T  I    ,  .      16  T  I 

end  and  equal  to  —  p  I2,  M  =  —p  I2  —  —  -  —  ;  hence  p  I  =  -  -  —  . 

8  O  fl  hi 

The  strength  is,  then,  f  times  as  great  as  for  the  same  load  in  the  middle, 
but  no  greater  than  for  a  beam  of  same  length  and  load  supported  at  both 
ends. 

18.  Beam  fixed  at  botli  euds—  Coii§tant  cros§-section— 
Concentrated  load.  —  Taking  our  notation  as  before  (Art.  12),  we 
have  in  this  case  not  only  a  reaction  at  the  right  end,  but  also  a  positive 
'moment  there  as  well,  both  of  which  must  be  found.  If  h  be  the  distance 
from  left  end  to  weight,  and  h  from  weight  to  right  end,  and  if  Vj  and 
V2  are  reactions,  Mi  and  M2  the  moments'  at  left  and  right  ends  respec- 
tively, then  for  equilibrium  we  must  have  —  Vi  +  V2  =  P,  Mj  +  Vi  ^  = 
M2-V3Za. 

For  x  less  than  ^  we  have  M  =  Mj  +V1x  =  EI  —  |.     Integrating 

cL  x 

once,  since  the  constant  is  zero,  because,  for  0  =  0,  ~  —  0,  we  have 

a  x 


d  x      2  E 
Integrating  again,  since,  for  x  =  0,  y  =  0,  and  the  constant  is  zero, 

x\ 


For  the  distance  from  the  right  end  to  the  weight  we  may  obtain  similar 
expressions,  if  we  take  that  end  as  the  origin,  only  we  should  have  —  V9 

and  M2  in  place  of  Vi  and  MI.     At  the  weight  itself  —  -    and  y  must  in 

(L  x 

each  case  be  equal,  but  *=-£  of  opposite  sign.     Therefore  we  have  the  equa- 

CL  $ 

tions  (2  MX  +  V,  Zx)  li  =  -(2  M2-V2  Z2)  Z2, 

(3  Mi  +  Vx  Zx)  Zx2  =  (3  M2-V2  Z2)  Za2. 


CHAP.  III.]  SUPPLEMENT   TO    CHAP.  VII.  123 

From  these  two  equations,  and  the  two  equations  above,  viz.,  ~  Vi  +  V2  =  P 
and  Mi  +  Vi  li  =  Mo—  V2  Z2,  we  can  determine  Vi,  V2,  Mi  and  M2. 
Thus  from  the  last  two  we  have 

Vi  Zi+V2  ?2  =  M2-M:  =V,  k  +  PZa  +  Vi  Z2=Vi  Z  +  FZ2, 

or  Vi  I  =  Ma-Mi-P  Z2. 

So  also  V2  Z  =  M2  -Mi  +P  Zi,  and  substituting  these  in  the  equations  above, 

we  have 

(Mi+Ma)  ZrrFZj  Z2, 

M!  Z  (2  ZX-Z2)-M2  Z  (2  Za-ZO  =  P  Zi  Z2  (k  -  Z2)  ; 
and  from  these  we  have,  finally, 


and  then  from  the  values  of  Vi  Z  and  V2  Z  above 


Vl  =  -P"2    v"^/     v^  = 


Z3  Z3 

Change  of  shape. 
Substituting  these  values,  we  can  now  find 


077 

Hence  y  is  a  maximum  for  x  = ^-,   and  the  maximum  deflection 

,        3  Zi  +  ly 

itself  is 

2  P  Zx3  Z22 


A  = 


3  E  I  (3 


This  expression  will  be  itself  a  maximum  for  Zi  =  Za  or  Zi  =  \  Z,  that  is, 
the  maximum  deflection  for  a  weight  in  the  middle  is  at  the  weight  and 
equal  to 

PZ3 


A  = 


192  E  I 


This  deflection  is  greater  than  the  maximum  deflection  for  any  other 
position  of  the  weight,  which  in  general  is  not  found  at  the  weight  itself, 
but  at  some  other  point  between  the  weight  and  farthest  end. 

We  see  above  that  the  deflection  in  this  case  for  load  in  middle  is  only 
one-fourth  as  much  as  for  same  beam  and  load  when  supported  at  the  ends. 

Breaking  weight. 

For  the  greatest  moment,  which  we  easily  find  to  be  at  the  end,  we  have 


_PZ1Z22_  PZ,  (Z-ZO 

~~          ~ 


This  is  a  maximum  for  Zi  =  £  Z.     That  is,  the  greatest  moment  at  the  end 
occurs  wlien  the  load  is  distant  one-third  of  the  length  from  that  end.    The 


124  SUPPLEMENT   TO    CHAP.  VII.  [CHAP.  in. 

value  of  this  greatest  moment  is  — -  P  1.     Hence  we  have  from  (11) P  I 

A  t  27 

2  T I  27  T I         27 

=  — 7 —  or  P  =  ,  or  —  as  great  as  for  the  same  beam  supported  at 

/i  d>  fi  I  lo 

p  7 

the  ends  only.     If  the  weight  is  in  the  middle,  however,  we  have  —  — 

8 

2  T  I  1  6  T  I 

— — -  or  P  =  — — — ,  or  twice  as  much  as  the  same  beam  supported  at  the 
tl  ii  L 

ends. 

19.  The  above  is  sufficient  to  introduce  the  reader  to  the  theory  of 
flexure.  He  can  now  discuss  for  himself  the  above  case  for  uniform  load, 
and  prove  that  the  maximum  deflection  is  at  the  centre  and  equal  to 

pi2  1 

.      That  the  greatest  moment  is  at  the  end  and  equal  to  —  p  l\ 

24  T   I 

and  that  the  breaking  weight  is  p  I  =  — .     We  may  also  observe  that 

ft   v 

both  in  the  beam  fixed  at  one  end  and  supported  at  the  other,  and  fixed  at 
both  ends,  the  moment  at  the  fixed  end  is  positive.  From  this  end  it  de- 
creases towards  the  weight,  and  finally  reaches  a  point  where  the  moment 
is  zero.  Past  this  point  the  moment  becomes  negative,  and  in  the  case  of 
the  beam,  free  at  the  other  end,  increases  gradually  to  a  maximum  and  then 
decreases  to  zero.  In  the  beam  fixed  at  both  ends,  it  increases  to  a  maxi- 
mum, then  decreases  to  zero,  then  changes  sign  and  becomes  positive  and 
increases  to  the  other  end.  These  points  at  which  the  moments  are  zero 
are  points  of  inflection,  because  here  the  curvature  changes  from  convex  to 
concave,  or  the  reverse. 

They  can  be  easily  found  from  the  equations  for  the  moments  by  finding 
the  value  of  x  necessary  to  make  the  moments  zero. 

Thus,  for  a  beam  fixed  at  one  end  and  supported  at  the  other,  uniform 

load,  the  inflection  point  is  at  a  distance  from  the  fixed  end  x  —  —  .      For 

both  ends  fixed,  we  make  M  —  —  p  [I*  —  6  (I  —  x)  x]  =  0,  and  find  x  = 

12 

V  (3  T  VW)  I  =  0.21131  I  and  0.7887  Z.  The  reader  will  also  do  well  to 
b 

discuss  the  curves  of  moments.  He  will  find  the  moments  represented  by 
the  ordinates  to  parabolas,  and  limited  by  straight  lines  similarly  to  Figs. 
73  and  75,  PI.  18. 

We  shall  give  in  the  Supplement  to  Chap.  XIV.  much  more  general 
formulae,  from  which,  for  one  or  both  ends  fixed  or  free,  the  moments  and 
reactions  at  the  supports  may  be  found,  when  any  number  of  spans  of  vary- 
ing length  intervene,  for  single  load  anywhere  upon  any  span,  or  uniformly 
distributed  over  any  span. 


CHAP.  VIII.]  CONTINUOUS    GIRDERS.  125 


CHAPTER    VIII. 


APPLICATION    OF   THE    GRAPHICAL    METHOD    TO    CONTINUOUS 
GIRDKRS GENERAL   PRINCIPLES. 

8O.  Mohr's  Principle. — Thus  far,  in  addition  to  the  general 
principles  of  the  Graphical  method,  we  have  noticed  more  or 
less  in  detail  its  application  to  the  composition  and  resolu- 
tion of  forces,  and  the  corresponding  determination  of  the 
strains  in  the  various  pieces  of  such  framed  structures  as  Bridge 
Girders,  Roof  Trusses,  etc.  We  have  also  illustrated  the 
graphical  determination  of  the  centre  of  gravity  and  moment 
of  inertia  of  areas,  as  also  of  the  bending  moments  and  shear- 
ing forces  for  simple  girders,  including  several  important  cases 
in  practical  mechanics.  (See  Art.  41.)  Lastly,  we  have  taken  up 
the  subject  of  Bridge  girders  more  in  detail,  and  developed  in 
order  the  principles  to  be  applied 'in  the  solution  of  any  par- 
ticular case.  Although  brief,  it  is  hoped  that  this  portion  will 
be  found  sufficient  to  illustrate  fully  the  method  of  procedure 
to  be  followed  in  practice. 

As  regards  simple  girders,  the  principles  referred  to  are  so 
easy  of  application  that  the  reader  will  find  no  difficulty  in 
diagraming  the  strains  in  any  structure  of  the  kind,  as  explained 
in  the  "  practical  applications  "  of  Arts.  8  to  13  ;  or  he  can  find 
the  maximum  moment  at  any  cross-section  for  given  load- 
ing according  to  the  last  chapter.  In  the  case  of  beams  or 
girders  continuoii,s  over  three  or  more  supports,  however,  we 
meet  with  difficulties  which  for  some  time  were  considered  in- 
superable. 

Thus  Cidinann,  in  the  work  which  we  have  so  often  quoted, 
says  :  *  "  The  determination  of  the  reactions  at  the  supports  for 
a  continuous  beam,  which  depend  upon  the  deflection,  the  law 
of  which  is  given  by  the  theory  of  the  elastic  line,  is  impossi- 
ble by  the  graphical  method,  at  least  so  far  as  at  present  de- 
veloped. The  theory  rests  upon  the  principle  that  the  radius  of 

*Culmann?8  GrapMsche8ta.Uk,  p.  278. 


126  CONTINUOUS    GIKDERS.  [CHAP.  VIII. 

curvature  of  the  deflected  beam,  for  any  cross-section,  is  in- 
versely proportional  to  the  moment  of  the  exterior  forces. 
'Now  the  deflection  at  any  point  is  so  small,  and  the  radius  of 
curvature  so  great,  that  its  construction  is  impracticable,  and 
will  so  remain  until  Geometry  furnishes  us  with  simple  rela- 
tions between  the  corresponding  radii  of  curvature  of  pro- 
jected figures  whose  projection  centre  lies  in  the  vertical  to  the 
horizontal  axis  of  the  beam.  If  such  relations  were  known,  we 
could  by  projection  exaggerate  the  deflection  of  the  beam 
until  the  radius  of  curvature  became  measurable.  Since  we 
are  not  yet  able  to  do  this,  we  must  have  recourse  to  calcula- 
tion" He  then  enters  into  a  somewhat  abstruse  analytic  dis- 
cussion of  the  continuous  girder,  and  deduces  formulae  for  the 
reactions  at  the  supports.  These  being  thus  known,  the  graph- 
ical method  is  then  applied. 

Concerning  this  difficulty,  Mohr*  remarks  that  it  has  but 
little  weight,  and  may  be  easily  overcome  if  the  same  simplifi- 
cation of  the  graphical  method  is  made  which  is  considered 
allowable  in  the  analytical  investigation,  viz.,  when  we  take  in- 


stead  of  the  exact  value  of  the  radius  of  curvative 


as  given  by  the  calculus,  the  approximate  value 


Thus,  let  PL  14,  Fig.  50  represent  a  perfectly  flexible  cord 
A  B  D  loaded  by  arbitrary  successive  forces.  The  variation  of 
these  forces  per  unit  of  horizontal  projection  dvwe  represent  by 
p.  Take  the  origin  of  co-ordinates  at  the  lowest  point  B.  If 
the  cord  is  supposed  cut  at  B  and  D,  we  have  at  B  a  horizontal 
force  H,  and  at  D  a  strain  S,  which  may  be  resolved  into  a 
horizontal  force  Hj  and  a  vertical  force  V.  Since  these  forces 
are  in  equilibrium  with  the  external  forces,  the  conditions  of 
equilibrium  are 

(1)  .......  H  =  H! 

and  px 

(2)  .....  V=   /     pdx. 

Jo 

*  Zeitsoh.  fas  kannov.  Arch,-u.  Ing.  Vereim.  —  Band  xiv.,  Heft  1. 


CHAP.  TUT.]  CONTINUOUS   GIRDERS.  127 

Moreover,  Cx 

I     p  dx 

(3) .      .  *La  = — =  — 

dx        H!  H 

Differentiating : 

d2  y      p  d  x 

«1    •  ± ryp 

~7 

6?    X 


Now,  had  we  formed  a  force  polygon  by  laying  off  the  forces, 
then  taken  a  pole  at  distance  H  and  drawn  lines  from  pole  to 
ends  of  forces,  the  corresponding  equilibrium  polygon  would, 
as  we  have  seen,  Art.  43,  be  tangent  to  the  curve  A  B  D  at  the 
points  midway  between  the  forces.  The  greater  the  number 
of  forces  taken,  the  shorter,  therefore,  the  sides  of  the  polygon  ; 
the  nearer  it  will  approach  the  curve  A  B  D.  This  curve  is 
therefore  the  equilibrium  curve,  found  according  to  the  graph- 
ical method.  Its  equation  is  given  above  by  (4). 

But  the  equation  of  the  elastic  line  is,  as  is  well  known, 

(y)  »•>'**•    7  'o        ~r~  ? 

a  (E>          J- 


where  E  is  the  modulus  of  elasticity  of  the  material,  M  the 
moment  of  the  exterior  forces,  and  I  the  moment  of  inertia  of 
the  cross -section. 

Comparing  now  this  equation  with  equation  (4)  above,  we 
see  that  the  elastic  line  is  an  equilibrium  curve  whose  horizon- 
tal force  H  is  E,  and  whose  vertical  load  per  unit  of  length  p 

is  represented  by  the  variable  quantity  -«-• 

This  simple  relation,  first  given  by  Mohr,  renders  possible 
the  graphical  representation  of  the  elastic  line,  and  not  only 
solves  graphically  almost  all  problems  connected  with  it,  but  in 
many  cases  simplifies  considerably  the  analytical  discussion 
also. 

81.  Elastic  Curve. — If  we  choose  the  pole  distance  H  at  -th 

n 

E  instead  of  E,  the  ordinates  of  the  elastic  line  will  be  n 
times  too  great.  If  the  scale  of  the  figure  is,  however,  -  th  the 

*  Stoney—  Theory  of  Strains,  p.  146.  Wood— Resistance  of  Materials,  p.  98. 
Also  Supplement  to  Chap.  VII ,  Art.  11. 


128  CONTINUOUS    GIRDERS.  [CHAP.  VIII. 

real  size,  then  in  the  diagram  the  ordinates  of  the  elastic  line 
will  l)e  given  in  true  size. 

Equation  (5)  may  also  be  written 


that  is,  the  elastic  curve  is  an  equilibrium  curve,  or  catenary, 
whose  horizontal  force  H  is  E  I  or  F,  and  whose  corresponding 

variable  load  per  imit  of  length  is  M  or  —  respectively. 

If  we  divide,  then,  the  moment  area  by  verticals  into  a  number 
of  smaller  areas,  and  consider  these  areas  as  forces  acting  at 
their  centres  of  gravity,  these  forces  determine,  as  we  have  seen 
(Art.  43),  an  equilibrium  polygon  which  is  tangent  to  the  elas- 
tic curve  at  the  verticals  which  separate  the  areas.  Thus  we 
can  construct  any  number  of  tangents  to  the  elastic  curve  ;  areas, 
which  are  positive  or  negative,  must,  of  course,  be  laid  off  in  the 
force  polygon  in  opposite  directions. 

If  we  divide  the  moment  area  by  lines  which  are  not  vertical 
[PL  14,  Fig.  51],  the  directions  of  the  outer  polygon  sides  are 
the  same  as  for  vertical  divisions,  because  the  vertical  height 
between  the  corresponding  outer  sides  in  the  force  polygon  is 
in  any  case  always  equal  to  the  total  load. 

The  two  outer  polygon  sides  for  any  method  of  division  are, 
therefore,  tangents  to  the  elastic  curve  at  the-  ends  of  the  same. 
Here  also  we  can,  of  course,  have  negative  areas. 

82.  Effect  of  End  Moments.—  A  beam  or  girder  continuous 
over  three  or  more  supports  differs  from  a  beam  simply  resting 
upon  its  supports,  in  that,  in  addition  to  the  outer  forces,  we 
have  acting  at  each  intermediate  support  a  moment  or  couple. 

But,  as  we  have  seen,  Art.  23,  the  effect  of  these  moments  or 
couples  will  be  simply  to  shift  the  closing  line  of  the  equilibrium 
polygon  through  a  certain  distance.  Thus  [PL  14,  Fig.  52  («)], 
if  the  span  ^  were  uniformly  loaded  and  simply  supported  at 
the  extremities  A  and  B,  the  equilibrium  curve,  or  curve  of  mo- 
ments, would,  as  we  know  (Art.  44),  be  a  parabola  A  D  B.  If, 
however,  the  beam  is  continuous,  we  have  at  A  and  B  moments 
or  couples  acting,  and  the  closing  line  A  B  is  shifted  to  some 
position  as  A'  B'.  If  now  wre  consider  the  moment  area,  we 
see  that  by  the  shifting  of  the  closing  line  the  former  moment 
area,  which  we  shall  call  \\\Q  positive  area,  is  diminished,  while 
to  the  right  and  left  we  have  negative  areas  A  A'  C  and  B  B'  C.' 


CHAP.  VIII.]  CONTINUOUS   GIRDERS.  129 

It  is  evident  that  these  areas  have  also  a  corresponding  action 
upon  the  elastic  line.  For  a  positive  moment  area  this  last  is 
concave  upwards,  while  for  negative  areas  it  is  convex  upwards. 
At  the  points  of  transition  C  and  C'  we  have  the  inflection 
points.  This  follows  easily  if  we  only  hold  fast  the  manner  in 
which  the  elastic  line  is  constructed,  viz.,  by  dividing  the  mo- 
ment area  into  laminae  and  regarding  the  area  of  each  as  a 
force.  The  forces  thus  obtained  must  plainly  act,  some  upwards 
and  some  downwards,  and  the  corresponding  equilibrium  poly- 
gon or  elastic  line  must  be  in  part  convex  upwards  and  in  part 
convex  downwards,  and  hence  at  the  points  of  transition  we 
must  hav re  points  of  inflection  where  the  moment  is  zero. 

83.  Division  of  tne  Moment  Area. — We  shall  assume  the 
cross-section  of  beam  constant.  Regarding  the  elastic  line 
simply  as  an  equilibrium  polygon,  we  can  apply  the  principle 
that  the  order  in  which  the  forces  are  taken  is  indifferent  (Art. 
6)  when  the  resultant  only  is  desired.  Since  in  the  considera- 
tion of  a  single  span  only  the  first  and  last  sides  are  of  impor- 
tance, we  can,  so  long  as  we  consider  a  single  span  only,  take 
then  the  laminae  or  divisions  of  the  moment  area  in  any  order 
we  please.  More  than  this,  we  can,  as  we  have  seen  in  Art.  81, 
divide  the  moment  area  into  laminae  not  vertical ;  for  example, 
we  may  in  any  span  distinguish  three  parts,  one  positive  and 
two  negative,  and  consider  each  as  a  force  acting  at  the  centre 
of  gravity  of  the  corresponding  area.  [This  holds  good  only 
for  constant  cross-section.  For  variable  cross-section  the  hori- 
zontal force  E  I  is  variable.]  Still  further,  we  can  divide  the 
moment  area  for  a  single  span  into  a  positive  area,  which  is  pre- 
cisely the  same  as  for  a  non-continuous  beam,  and  into  a  nega- 
tive area,  which  will  be  evidently  a  trapezoid. 

This  is  of  great  importance.  To  understand  it  fully  we  refer 
to  PL  14,  Fig.  52.  Here,  in  the  second  span,  we  see  that  the 
real  moment  area  consists  of  a  positive  part,  viz.,  the  parabola 
C  D  C',  and  two  negative  parts  A  A'  C  and  B  B'  C'.  Instead  of 
these  we  may  take  the  entire  parabolic  area  A  D  B  and  the  tra- 
pezoid A  A'  B'  B,  or,  finally,  instead  of  this  trapezoid,  we  may 
take  the  two  triangles  A  A'  B'  and  B  B'  A'.  The  parabolic 
area  is  positive,  the  triangular  areas  are  negative. 

If  we  assume  the  load  as  uniformly  distributed,  the  first  area 
will  be  always  parabolic,  and  we  may,  therefore,  call  it  the 
parabolic  area. 


130  CONTINUOUS    GIRDERS.  [ciIAr.  Vm. 

By  this  division  of  the  moment  area  we  have  obtained  a  great 
advantage.  While  the  three  areas  C  D  C',  A  A'  C  and  B  B'  C' 
are  all  three  dependent  upon  the  moments  at  the  supports  A  A! 
and  B  B',  we  have  by  £his  new  division  to  do  with  three  areas, 
of  which  the  first  is  entirely  independent  of  the  moments  at  the 
supports,  the  second  depends  only  upon  that  to  the  left,  and  the 
third  only  upon  that  to  the  right. 

84.  Properties  of  the  Equilibrium  Polygon. — Let  us  con- 
sider now  the  case  of  a  beam  over  four  supports,  that  is,  of 
three  spans — Z0,  4  and  4 — the  first  and  last  being,  as  is  usually  the 
case,  equal,  and  the  two  first  loaded  with  both  live  and  dead 
load,  the  last  with  dead  load  only.  The  parabolas  for  the  ver- 
tical loads  [PL  14,  Fig.  52]  may  be  constructed  by  means  of  a 
force  polygon,  or  the  ordinates  at  the  centre  calculated,  and  the 
parabolas  then  drawn.  The  moments  at  the  supports  are  A  A' 
and  B  B'.  Although  these  are  unknown,  it  is  not  necessary  to 
assume  them  at  first.  They  may  be  directly  constructed. 

Thus,  if  we  conceive  the  moment  areas  in  each  span  divided 
into  positive  parabolic  areas  and  negative  triangles,  we  have  in 
the  first  and  last  span  one,  in  the  middle  two  triangles.  If  we 
consider  these  areas  as  forces  acting  at  the  corresponding  centres 
of  gravity,  we  shall  obtain  an  equilibrium  polygon  of  the  form 
given  in  Fig.  52  (b).  That  is,  this  polygon  must  have  eight 
sides,  and  its  angles  must  be  somewhere  on  the  verticals  through 
the  centres  of  gravity  of  the  parabolic  and  triangular  areas. 
The  parabolic  areas  act  downwards,  the  triangular  areas  up- 
wards. The  problem  is,  to  make  these  last  so  great  that  this 
polygon  shall  pass  through  all  the  points  of  support. 

One  of  the  properties  of  the  polygon  we  have,  therefore,  just 
noticed,  viz. :  its  angles  must  lie  in  the  verticals  through  the 
middle  points  of  the  spans  and  through  the  points  distant  from 
A  and  B  one-third  of  the  spans  on  each  side  (i.e.,  the  centres  of 
gravity  of  the  triangles). 

If  we  prolong  the  second  and  fourth  sides  of  the  polygon, 
they  intersect  in  a  point  M,  the  point  of  application  of  the 
resultant  of  the  two  contiguous  triangular  area  forces  (Art.  44). 

The  areas  of  these  two  triangles  are  -  A  A'  Z0  and  -  A  A'  ^,  that 

is,  the  areas  are  as  the  spans  4  and  ^. 

Then  by  the  principle  of  Art.  18  the  resultant  divides  the 
distance  between  the  forces  into  two  portions,  which  are  to  each 


CHAP.  VHI.]  CONTINUOUS    GIRDERS.  131 

other  as  ^  to  4> or  inversely  as  the  forces.     Since  the  entire  dis- 
tance is  -  £0+  7:  ^i>  the  distance  of  the  resultant  or  of  the  point 
o          o 

of  intersection  M  from  L  is  -  ^ ;  from  N  it  is  -  1Q.     The  point 

O  t> 

M,  therefore,  must  lie  somewhere  in  the  vertical  at  -  ^  from  L, 

o 

the  point  of  application  of  the  triangular  area  force  for  the 
span  4-  The  verticals  through  the  centres  of  the  parabolic 
areas  we  call  the  parabolic  or  middle  verticals  /  those  through 
the  centres  of  gravity  of  the  triangular  areas,  the  third  verti- 
cals /  those  through  a  point  as  M,  the  point  of  application  of 
the  resultant  of  two  contiguous  triangular  area  forces,  the  lim- 
ited third  verticals.  Upon  these  verticals  two  sides  must  always 
intersect. 

85.  Polygon  for  tlie  Positive  Moment  Areas. — It  will  be 

found  best  to  take  as  the  reduction  base  for  areas  —  ^,  i.e.,  half 

2 

the  second  span,  and  for  pole  distance  —  l±.      Reducing  the 

3 

areas  of  the  parabolas  to  this  basis,  and  considering  the  heights 
thus  obtained  as  forces,  we  can  form  a  force  polygon  with  pole 

distance  -  \.     It  is  not  necessary  to  draw  this  polygon ;  our  ob- 
3 

ject  is  to  find  the  corresponding  equilibrium  polygon.  This 
last,  since  we  consider  the  entire  parabolic  area  as  a'force  act- 
ing at  its  centre  of  gravity,  consists  of  two  lines  which  inter- 
sect in  the  vertical  through  the  middle  of  the  span.  We  pro- 
long each  of  these  lines  and  obtain  two  lines  as  shown  in  PL  14, 
Fig.  52  (c).  The  segments  cut  off  by  these  lines  from  the  verti- 
cals through  the  supports  are  the  moments  of  the  parabolic 
areas  with  respect  to  the  supports.  These  moments  we  can 
easily  find. 

Thus,  let  the  deflection  of  the  parabola  in  the  second  span  at 

o 

the  centre  be  f,  then  its  area  is  —  fl^.      This   reduced   to   the 

3 

1  4: 

basis  -  4  gives  -/  as  the  force.     The  moment  of  this  force 
A  o 

4:  1  1 

with  reference  to  the  supports  is  —  f  x  —li  =  2fx—li.     This 

32  3 

moment  is  equal  to  the  segment  sought  multiplied  by  the  pole 


132  CONTINUOUS    GIRDERS.  [CHAP.  VIII. 

distance.     This  last  is  -  ^.    The  segment,  therefore,  is  2/!    "We 

o 

do  not  need,  therefore,  to  draw  the  force  polygon,  but  have  sim- 
ply to  take  off  with  the  dividers  the  middle  or  din  ate  of  the  para- 

/D  $ 

bola  f  =  ~rr ,  and  lay  it  off  twice  on  the  verticals  right  and  left 

8 

through  the  supports,  and  join  the  four  points  thus  obtained  by 
lines  crossing  each  other  under  the  centre  of  the  span.  The 
equilibrium  polygon  for  the  positive  parabolic  area  is  then 
ready  for  the  middle  span.  ["We  advise  the  reader  to  construct 
it  for  himself.] 

For  the  two  side  spans  the  construction  is  different.     Here 

2  4       I 

the  area  of  the  parabola  is  -f  10)  the  reduced  area  -f  j,  and 

3  3       L\ 

the  moment  x     =  2'       . 


Dividing  by  -  ^,  we  have  for  the  segment  required 
3 


Therefore,  in  the  end  spans,  or  generally  in  any  span  not  equal 

to  the  standard  span,  or  that  which  furnishes  the  constants  -  l± 

2 

and  -  Zj,  we  must  multiply  the  middle  ordinate  of  the  parabola 
o 

by  the  square  of  the  ratio  of  the  "  standard  n  span  to  the  span 
in  question,  and  then  lay  the  product  off  twice  upon  the  verti- 
cals through  the  supports.  This  multiplication  is  easily  per- 
formed graphically.  If  from  the  middle  of  span  4  we  lay  off 
If  horizontally  and  join  the  end  with  the  end  of  y,  then  lay  off 
42  in  same  direction  and  draw  a  parallel  to  the  first  line,  the 

lz 
segment  on/"7  will  bejf'~&     For  we  shall  have 

h 

79  /» /    •    •      V  O  /*/   ''0 

42:/  ::  tf:^    or  os=f 'A. 

Since  these  cross-lines  depend  upon  given  quantities,  they 
can  be  constructed  for  every  span,  and  thus  we  have  Fig.  52,  c. 
They  give  us  not  only  the  moments  over  the  supports,  but  also 
the  moments  for  any  point  of  the  parabolic  area  forces.  We 
shall  hereafter  make  use  of  them. 

$6.  Construction  of  the  Fixed  Points,  and  of  the  Equi- 
librium Polygon. — We  have  thus  all  the  given  and  known 


CHAP.  Vni.]  CONTINUOUS    GIRDERS.  133 

quantities,  have  deduced  the  general  properties  of  the  equili- 
brium polygon,  and  will  now  endeavor  by  their  aid  to  draw  the 
polygon  itself.  We  shall  then  be  able  to  find  the  actual  mo- 
ments A  A'  and  B  B'  at  the  supports.  It  is  impossible  at  first 
to  draw  any  single  side  of  the  polygon  in  true  position,  and  we 
must,  therefore,  endeavor  to  find  certain  /points  of  the  same  suf- 
ficient to  determine  it. 

Lay  off  first  the  three  spans,  PL  14,  Fig.  52  (d).  Suppose 
the  second  side  of  the  polygon  prolonged  till  it  intersects  the 
vertical  through  the  end  support  a,  in  a  point  K,  Fig.  52  (£). 
This  point  is  known.  It  is  given  by  the  moment  of  the  para- 
bolic area  in  the  first  span  with  respect  to  this  end  support. 
This  moment  we  have  already  by  the  cross-lines  in  Fig.  52  (c). 
We  have  then  simply  to  take  it  off  in  the  dividers  from  (c)  and 
lay  it  off  from  d  to  K7  in  Fig.  52  (d).  We  have  now  in  Fig.  52 
(J),  two  points  of  the  polygon  known,  namely,  the  end  support 
and  K,  which  last  must  be  in  the  second  side  prolonged. 

The  triangle  L  M  N  is  now  of  special  importance.  What- 
ever may  be  the  position  of  KM  and  M  N,  we  have  already 
seen  that  the  intersection  M  must  always  lie  somewhere  in  the 
limited  third  vertical.  The  first  side  KM  must,  however ', 
always  pass  through  K,  a  known  point.  The  second  must  pass 
through  the  support,  also  a  known  point.  The  points  L  and  N 
must,  moreover,  always  lie  in  the  third  verticals,  distant  from 

A,  5-  \  and  ^  Z0  respectively. 

If  the  line  K  M  takes  up  various  positions  under  these  con- 
ditions, the  line  M  N  will  revolve  about  a  fixed  point  which  is 
given  ~by  the  intersection  of  a  line  through  K  and  the  support 
A  with  M  N. 

If,  then  [Fig.  52  (d)],  we  draw  a  line  in  any  arbitrary  direction 
through  K7,  and  note  the  intersections  L'  and  M7  with  the  first 
third  vertical  and  the  limited  thtrd,  then  through  L'  and  the 
support  draw  a  line  to  intersection  N7  with  second  third  verti- 
cal, and  join  M7  N7,  and  finally  through  K'  and  the  support 
draw  a  line  intersecting  this  last  in  I,  the  point  I  thus  deter- 
mined is  &  fixed  point,  and  remains  the  same  for  any  position 
of  K'  M7.  It  is  therefore  a  point  on  the  fourth  side  M  N  of 
the  polygon.  For  the  triangle  L7  M'  N'  may  have  any  posi- 
tion, yet  so  long  as  its  angles  lie  in  three  parallel  fixed  lines, 
ui id  two  of  the  sides-  pass  through  two  fixed  points,  the  other 


134  CONTINUOUS    GIRDERS.  [CHAP.  VIII. 

side  must  also  pass  through  a  fixed  point.*  Out  of  all  possible 
positions  of  the  triangle,  one  of  these  positions  must  coincide 
\vith  the  polygon  sides,  and  hence  this  fixed  point  is  a  third 
known  point,  since  we  have  already  K'  and  the  end  support. 

Although,  then,  we  are  as  yet  unable  to  draw  any  of  the  sides 
of  the  polygon  in  true  position  and  direction,  still  from  the 
hitherto  known  properties  we  have  deduced  a  new  one.  We 
know  now  a  point  through  which  the  fourth  side  must  pass. 
But  this  is  not  all.  We  proceed  still  further.  The  fourth  and 
fifth  sides  must  intersect  upon  the  vertical  through  the  centre 
of  the  second  span.  These  sides,  moreover,  cut  off  upon  any 
vertical  the  moment  of  the  parabolic  area  with  respect  to  any 
point  in  that  vertical.  We  know  this  moment  thus  for  the 
point  I  just  found.  It  is  found  by  taking  the  segment  cut  off, 
from  the  vertical  through  I,  by  the  cross-lines  for  the  parabolic 
areas  found  above  in  Fig.  52  (c). 

Laying  this  segment  off  from  I,  we  thus  find  I',  a  point  in  the 
fifth  side  prolonged.  From  this  point  we  proceed  as  before  to 
find  the  next  fixed  point  I".  We  then  lay  off  from  1"  the  mo- 
ment of  the  parabolic  area  for  this  point  and  find  F",  a  point 
upon  the  eighth  side.  We  can  now  draw  the  polygon  itself. 

Thus  the  eighth  side  passes,  of  course,  through  the  last  sup- 
port and  also  I"'.  It  is  therefore  determined.  Through  the 
intersection  of  this  line  with  the  vertical  through  the  middle  of 
the  span  and  the  point  I"  the  seventh  side  passes.  The  sev- 
enth side  is  therefore  determined.  Through  the  intersection  of 
this  with  the  third  vertical  and  the  support  the  sixth  side 
passes  and  continues  till  it  intersects  the  third  vertical  on  the 
other  side.  Then  from  this  point  towards  I'  to  intersection 
with  vertical  through  centre  of  middle  span.  From  this  last 
point  towards  I  to  intersection  with  third  vertical.  From  this 
last  point  again  through  support  to  intersection  with  third  verti- 
cal on  other  side  ;  then  towards  K'  to  intersection  with  vertical 
through  centre  of  end  span ;  and  lastly,  from  this  last  point  of 
intersection  through  the  end  support,  and  the  polygon  is  com- 
plete as  given  in  Fig.  52  (b). 

In  this  manner,  however,  inaccuracies  may  occur.  To  avoid 
these  we  may  start  from  the  rig/it  end  support  and  also  find 
four  fixed  points  as  above.  It  is  unnecessary  to  make  the  con- 

*  This  proposition  the  reader  can  easily  prove  geometrically  or  analytically. 
See  Art.  112. 


CHAP.  VIII.]  CONTINUOUS    GIRDEKS.  135 

striiction.  We  see  at  once  that  we  shall  thus  obtain  in  each  end 
span  two  points,  in  the  middle  span  four  points,  which  last,  be- 
ing joined  by  lines  crossing  each  other,  give  in  the  middle  span 
two  sides  in  proper  position.  It  is  also  evident  how  the  poly- 
gon may  then  be  completed. 

87.  Construction  of  the  Moments  at  the  Support.*. — Thus 
we  are  able  to  construct  the  equilibrium  polygon,  or  rather  the 
extreme  tangents  to  the  elastic  line  for  each  span.  "We  have 
now  to  determine  the  two  moments  over  the  supports.  This  is 
very  simple.  The  first  moment  to  the  left  is  cut  off  by  the 
fourth  side,  the  second  by  the  fifth  side  of  the  polygon,  from 
the  verticals  through  the  supports.  We  have  therefore  only  to 
prolong  these  two  sides,  take  off  the  segments  in  the  dividers, 
and  lay  them  off  in  Fig.  52  (a)  in  A  A!  and  B  B'.  We  have, 
then,  in  PL  14,  Fig.  52  (a)  the  moments  for  the  given  case  and 
loading  at  any  point,  as  shown  by  the  shaded  area. 

The  proof  is  simple.  The  two  lines  N  M  and  N  L  [Fig.  52, 
Z*]  evidently  cut  off  upon  the  vertical  at  the  support  the  moment 
of  the  force  acting  at  N.  This  force  is  the  area  of  the  triangle 

A  A.'  B'  equal  to  -  A.  A!  ^.    This  reduced  to  the  basis  -  l±  gives 

A  A!.  If.  we  multiply  this  by  the  lever  arm  of  the  force,  we 
have  its  moment.  This  moment  is,  however,  equal  to  the  seg- 
ment A  A'  multiplied  by  the  pole  distance,  and  since  this  pole 

distance  is  itself  ~  h  the  segment  itself  to  the  assumed  pole 

o 

distance  gives  us  the  moment. 

We  see  that  it  is  not  necessary  to  draw  the  line  N  L  as  it 
passes  through  the  support.  We  have  simply  to  prolong  the 
side  M  N  to  intersection  with  the  vertical  through  the  support. 
It  is  to  be  observed  that  the  moments  at  the  supports  are  cut  off 
at  the  supports  only  by  those  lines  which  pertain  to  the  "  stand- 
ard "  span,  or  that  span  from  which  we  take  our  reduction  basis 
and  pole  distance.  For  lines  in  the  other  spans  the  above  does 
not  hold  good  without  modification.  It  is,  however,  always 
possible,  at  least  for  from  two  to  five  symmetrical  spans,  to 
observe  the  above  conditions.  In  those  cases  where  this  is  not 
possible,  an  easy  graphical  multiplication  of  the  segments  by 
the  square  of  the  ratio  of  the  spans  will  give  the  moments. 
We  see  also  the  reason  why,  for  four  symmetrical  spans,  the 
and  not  the  first  must  be  taken  as  the  standard  span. 


136  CONTINUOUS   GIBBERS.  [CHAP.  VIII. 

If  the  construction  of  the  moments  over  the  supports  is  our 
sole  purpose,  as  is  in  practice  the  case,  the  polygon  need  not  be 
drawn.  We  have  only  to  find  our  fixed  points,  and  note  the 
intersection  of  the  sides  with  the  verticals  through  the  supports, 
without  drawing  the  sides  themselves.  In  the  preceding  Arts- 
we  have  purposely  considered  only  the  particular  case  of  uni- 
form loading,  and  have  taken  only  three  spans,  in  order  to 
familiarize  the  reader  with  the  nature  of  the  problem  and  the 
method  of  its  solution.  In  oi'der  to  attain  a  clear  understand- 
ing of  the  subject  as  thus  far  developed,  he  would  do  well  to 
take  some  particular  case,  as,  for  instance,  that  of  a  girder  of 
two  or  three  or  four  spans  of  given  length,  the  end  spans  being 
equal,  and  intermediate  spans  equal  and  say  one-fourth  longer 
than  the  ends,  and  work  out  by  diagram  the  moments  at  the 
supports  for  a  uniform  load  over  the  whole  length  of  girder. 
For  two  spans  the  moment  at  the  centre  support  should  be 

g  p  P,  I  being  the  length  of  span,^>  the  load  per  unit  of  length. 

For  three  spans  the  moment   at  the  two   inner  supports   is 

1  -f-  n3 
4  (3  i  o — \ P  ft,  where  n  I  =  the  length  of  end  spans.     Thus,  if 

3  91 

n  =  -,  we  have  TT^Q  &  ^'     "^or  ^our  sPans  ^e  moment  at  the 

1-4-2  w3 
second  and  fourth  supports  is  .    Q       , — rp  P,  and  at  the  middle 

4:  \O  ~r  4:  7l>\ 

1  _l_  2  n nz 

support  4/014 — r  P  P-      By  these  formulae   the  graphical 

results  may  be  checked. 

When  the  reader  has  thus  become  thoroughly  familiar  with 
the  principles  of  the  preceding  Arts,  and  their  practical  appli- 
cation, he  will  be  ready  to  resume  at  this  point  the  more  gen- 
eral development  which  follows. 

88.  The  Second  Equilibrium  Polygon. — We  see,  there- 
fore, that  the  actual  form  of  the  elastic  line  is  not  required  to 
be  known.  Only  the  outer  forces  and  their  moments  are  sought, 
and  to  determine  these  it  is  sufficient  to  know  the  position  of 
the  tangents  to  the  elastic  line  at  the  supports.  Thus  the  first 
line  of  the  equilibrium  polygon  [Fig.  52,  &]  being  given  in 
position,  by  the  aid  of  the  middle,  third,  and  limited  third 
verticals  and  the  known  point  K,  all  the  other  sides  may  be 


CHAP.  VIII.]  CONTINUOUS    GIKDEKS.  137 

drawn,  and  the  moments  at  the  supports  found.  We  conceive, 
therefore,  the  moment  area  as  the  difference  of  the  trapezoid 
A'  A"  B"  B'  [PI.  15,  Fig.  53]  and  the  parabolic  area  A"  C"  B", 
or  equal  to  A"  C"  B"  minus  triang.  A7  A"  B'  'minus  triang. 
B/  B//  A//_  rp}ie  area  A//  c//  B//  we  cau  the  wrwpiQ  or  parabolic 

moment  area. 

If  we  indicate  the  moments  at  the  supports  A7  A."  and  B'  B" 
by  M'  and  M",  then,  for  a  given  span  Z, 

A'  A"-  B7  =  J  M'  Z  and 
A"  B'  B"  =  J  M"  I. 

If  we  indicate  further  the  height  of  a  rectangle  of  base  Z  and 
area  A7/  C"  B",  that  is,  the  mean  value  of  the  'moments  of  the 
corresponding  simple  girder  by  9)2,  we  have 

area  A"  C"  B"  =  3tf  Z. 

The  verticals  through  the  centres  of  gravity  of  the  triangles 
divide  the  span  into  three  equal  parts.  We  call  these  the  third 
verticals.  The  load  9ft  I  acts  at  the  centre  of  gravity  of  the 
parabolic  area  A"  B"  C". 

The  four-sided  equilibrium  polygon  A  U  S  V  B  corresponding 
to  these  forces  we  call  the  second  equilibrium  polygon.  The 

pole  distance  must  be  (Art.  81)  -El.    Instead  of  this,  we  take, 

itf 

which  amounts  to  the  same  thing,  the  forces  G  F  =  -  M'  - , 
EH—-  M"-  and  F  E  =  9ft  -,  and  the  pole  distance  1}  — 

2i  A»  A 

E  I 

— -,  where  \  is  any  assumed  length.  For  \  we  may  take  the 
n  A/ 

arithmetical  mean  of  all  the  spans,  or,  as  we  have  seen,  one  of 
the  actual  spans.  If  the  outer  spans  are  both  equal  to  10  and  the 
other  spans  equal  to  11?  we  should  naturally  choose  \  =  ?1?  since 

then  the  forces  would  be  -  M',  -  M"  and  9ft. 

2t  2t 

If  the  position  of  the  tangents  at  the  supports  were  known 
or  found,  the  equilibrium  polygon  could  be  easily  drawn  as 
follows :  Upon  the  two  verticals  distant  each  side  of  the  centre 

E  I 

S  [Fig.  53]  by  the  pole  distance  ~b  —  — -  lay  off  the   distance 

n  A» 


138  CONTINUOUS    GIBBERS.  [CHAP.   VIII. 

A  —  %)l  -  and  join  the  points  thus  obtained  by  two  lines  cross- 

ing each  other.  These  cross-lines  are  the  lines  OP  O  E  of  the 
force  polygon.  If  now  we  make  U  IT  and  V  V  equal  to  the 
ordinates  of  the  cross-lines  vertically  under  U  and  V,  then  the 
sides  of  the  equilibrium  polygon  U  S  and  V  Q  prolonged,  pass 
through  U'  and  V'.  This  will  at  once  appear  from  an  inspec- 
tion of  Fig.  53. 

In  this  form  the  equilibrium  polygon  was  first  repre- 
sented by  Mohr.  (Zeitschrift  des  Arch,  und  Ing.  Ver.  zu 
Hannover,  1868.) 

89.  Determination  of  the  Moments  over  tlie  §upports. 
—  If  we  draw  in  the  force  polygon,  lines  parallel  to  the  four 
sides  of  the  second  equilibrium  polygon,  then  the  segments  of 
the  force  line  between  the  lines  parallel  to  A  U,  B  V  [PI.  15, 
Fig.  53]  and  those  parallel  to  S  U,  S  V,  are  respectively  F  G  = 

-  M'  -  and  E  H  =  -  M"  -..  If  we  prolong  S  U  and  S  V  to 
A  A/  2t  A 

intersection^  M  and  N  with  verticals  through  the  supports,  and 
represent  A  M  and  B  N  by  y'  and  y",  we  have  from  the  simi- 
larity of  the  triangles  U  A  M  and  V  B  N  with  O  G  F  and  O  H  E 


hence 

'  -  ?^L£    "  -  M"^ 

y  ~  6  I  A    y     ~  6  I  A* 

The  segments  A  M  and  B  N  are,  therefore,  proportional  to 
the  moments  at  the  supports  Mx  and  M". 

These  moments  themselves  can  now  be  determined  in  vari- 
ous ways. 

\st.  It  is  in  general  best  to  choose  the  second  pole  distance 

b  =  -  \.     We  have  then 
o 

M'  =  y'  (          M"  =  y"  (Art.  70). 


If,  then,  at  a  distance  from  U  and  V  either  way  equal  to 

2bl-\  =  --\  we  draw  verticals,  the  segments  A!  M!  and 
\  LI       06 

Bj  Nj  cut  off  from  these  verticals  will  evidently  be  equal  to 
the  moments  required,  viz.,  1VT  and  M". 


CHAP.  VIII.].  CONTINUOUS    GIRDERS.  139 

%d.  If  we  take  X  =  I,  we  have  at  once  M/  =  y'  and  M"  =  y"  . 
In  the  inner  spans,  therefore,  we  have  directly,  as  we  have  al- 
ready seen,  for  the  special  case  (Art.  72),  when  \—l,  the  mo- 
ments at  the  supports. 

3d.  For  a  span  adjoining  a  span  whose  length  is  X,  we  have 
the  moment  for  the  intervening  support  directly  from  this  last 
span.  If  the  inner  spans  have  the  same  length  I,  and  the  two 
outer  the  same  length  £1?  we  can  accordingly,  by  making  \  =  I, 
obtain  directly  the  moments  at  the  supports. 

90.  Comparison  with  Girder  fixed  horizontally  at  both 
ends.  —  If  the  ends  are  fixed  horizontally,  the  lines  in  the  force 
polygon  parallel  to  A  U  and  B  V  coincide,  i.e.,  H  and  G  fall 
together.  Accordingly,  if  we  designate  the  end  moments  now 
by  W  W"  we  have  (Fig.  53) 

")  =  W. 

Therefore,  in  the  first  equilibrium  polygon,  the  moment  areas 
on  each  side  of  the  dosing  line  are  equal. 

Indicating  the  points  for  this  case  by  the  index  0  [Fig.  54, 
PL  15],  we  have  the  triangle  A  U0  M0  equal  to  U0  V0  V'o, 
and  therefore  A  M0  =  V0  V0  —  V  V,  as  also,  in  like  manner, 

B  N0  =  UolTo  =  U  U',  or,  taking   b  =  1  X, 

U  U'  =  3JT  /-\2      V  V  = 

Therefore,  the  ordinates  between  the  cross-lines  at  the  verti- 
cals passing  through  U  and  V  are,  for  girders  fixed  horizon- 
tally, proportional  to  the  end  moments  Wl'  and  9ft". 

For  X  =  I,   fi  =  -  I,  and  these  ordinates  give  the  moments 
6 

directly. 

If  we  draw  through  U7  and  V  a  straight  line  intersecting 
the  end  verticals  in  Q  and  R,  and  prolong  N  U'  and  N  V  to 
intersections  S  and  T,  then  Q  M  —  2  U  U7,  Q  S  =  V  V',  and 

hence  M  S  =  2  U  IF+V  V'  =  (2  2U'+3B")  (^Y;  and  in  the 


same  way  N  T  =  (2  2tt"  +2ft')  {-j\ 


Therefore,  the  segments  cut  off  upon  the  end  verticals  ly  the 
cross-lines  are  proportional  to2W+  W  and  2  2ft"  +  W. 


140  CONTINUOUS    GIEDEES.  [CHAP.  VIII. 

The  quantities  SDc'  and  $31"  being  known,  we  can  easily  con- 
struct the  cross- lines. 

If  we  draw  a  line  through  U  and  V  to  intersections  O  and  P, 
we  have  O  M  =  V  V,  P  N  ==  U  U'.  Therefore,  A  O  and  B  P 

are  equal  to  (>Itt'-M')  ^  and  (2JT-M")  f -)*,  where  M'  and 

M"  are  the  moments  at  the  supports  for  a  continuous  girder, 
and  9ft'  W  those  for  a  girder  horizontally  fixed  at  its  ends. 


CHAP.  IX.]       LOADED  AND  UNLOADED  SPANS.  141 


CHAPTER  IX. 

CONTINUOUS    GIRDER — LOADED    AND    UNLOADED    SPANS. 

91.  Unloaded  Span. — If  the  span  is  unloaded,  we  have  to 
construct  the  second  equilibrium  polygon,  only  the  two  forces 

-  M7  I  and  -  M"  I.   If  the  position  of  the  end  tangents  is  known, 

the  polygon  is  completely  determined.  If  we  prolong  the 
middle  side  U  V  to  intersections.  M  and  N  with  the  end  ver- 
ticals [PL  15,  Fig.  55],  then,  by  the  preceding  Art,  A  M  == 


MM       ,  B  N  =  M"    - )  ;    therefore,  A  M  :  B  N  ;  :  M/  :  M". 

\A//  \A// 

If  now  w^e  draw  A  B  intersecting  U  V  in  I,  the  moment  at  this 
point  is  zero.  That  is,  the  intersection  I  of  the  line  joining  the 
supports  with  the  middle  side  of  the  polygon  is  the  point  of  in- 
flection of  the  elastic  line. 

92.  Two  §ucce§§ive  Unloaded  Spans. — Prolong  the  two. 
middle  sides  U  V  and  Ut  Vt  [PI.  15,  Fig.  56]  of  the  equilibrium 
polygon  for  the  two  spans  Z0  and  ^.     The  point  of  intersection 
W  is  a  point  in  the  resultant  of  the  forces  at  V  and  U^     Since 

these  forces  are  -  JV^  1G  and  -  M:  ^,  we  have  W0  V0  :  W0  U0  : : 

li  :  4-      But  the  horizontal  projection  of  V0  U0  is  -  (70  +  k), 

3 

therefore  that  of  V0  W0  is  -  ^  and  of  U0  W0,  -  4 ;  while  that  of 

o  3 

B  W0  is  -  (li—lo).     The  vertical  through  W  we  have  called  the 
o 

limited  third  vertical.  Its  position  is,  as  we  see,  easily  found, 
and  depends  simply  upon  the  length  of  the  spans. 

Let  us  now  consider  more  closely  the  intersections  I  and  L  of 
the  middle  sides  with  the  straight  line  joining  the  supports. 
We  have 


14:2  CONTINUOUS    GIEDEK.  [CHAP.  IX. 

But  we  have  also 

UoU^VoV-BUoiBVo::  l,:l^ 
hence 

L  U0  :  L  W0 ; ;  I  V0  X  \  :  I  W0  x  10. 

The  ratio  of  the  parts  into  which  W0  U0  is  divided  by  the 
point  L  depends,  therefore,  for  given  spans  only  upon  the  ratio 
of  I  V0  to  I  W0,  or  upon  the  position  of  I  alone. 

If,  therefore,  we  were  to  draw  through  the  point  I  another 
polygon,  the  point  L  would  be  unchanged,  or  still  more  gener- 
ally, if  the  point  I  for  different  heights  of  the  supports  and 
different  polygons  moves  in  the  same  vertical^  the  point  L  will 
also  move  in  a  vertical. 

If  the  supports  are  in  the  same  straight  line,  the  points  I  and 
L  are  the  points  of  inflection  of  the  elastic  line.  We  have 
therefore  the  principle,  that  if  for  different  polygons  the  in- 
flection point  I  remains  the  same,  the  inflection  point  L  re- 
mains also  the  same. 

The  point  I  being  given,  we  can  easily  construct  the  point  L. 
We  have  only  to  draw  through  I  at  will  any  line  intersecting 
the  third  vertical  through  V0  and  the  limited  third  at,  say,  V 
and  W.  Through  V  and  the  support  B  draw  a  line  to  inter- 
section Ui  with  third  vertical  through  U0.  Join  now  U^  with 
W.  The  line  U^  W  cuts  the  line  through  the  supports  A  and 
B  in  the  point  L.  (See  also  Art.  86,  Fig.  52,  d.) 

93.  The  "Fixed  Points."— Suppose  that,  starting  from  the 
left  support  A  [PI.  15,  Fig.  57],  we  have  a  number  of  unloaded 
spans.  The  end  A  then  is  an  inflection  point,  since  the  mo- 
ment there  is  zero.  Starting  from  this  point,  therefore,  we  can 
construct,  according  to  the  preceding  Art.,  the  inflection  point  I2 
for  the  next  span.  Then  starting  from  this  we  may  construct 
the  point  I3  for  the  third  span,  and  so  on.  Since  these  points, 
under  the  assumption  that  the  supports  all  lie  in  the  same 
straight  line,  do  not  change  their  position,  whatever  may  be  the 
loading  of  the  loaded  spans,  and  whatever  spans  be  loaded,  we 
call  tlwm  fixed  points. 

A  second  series  of  fixed  points  may  be  in  similar  manner 
constructed,  when  a  number  of  spans  from  the  right  are  un- 
loaded, so  that  there  are  two  series  of  fixed  points.  In  the 
end  spans  the  end  supports  are  fixed  points. 

It  follows  directly  from  the  construction  that  the  fixed  points 
are  always  within  the  outer  third  of  the  span. 


CHAP.  IX.]        LOADED  AND  UNLOADED  SPANS.  143 

The  construction  of  the  fixed  points  is  the  first  operation  in 
the  graphical  treatment  of  the  continuous  girder. 

The  above  construction  was  first  given  by  Mohr. 

91.  Shearing  Force,  Reactions  at  the  Supports,  and  UIo- 
meitts  in  the  Unloaded  Spans. — The  moments  in  the  unloaded 
spans  are  given,  then,  by  the  ordinates  to  a  broken  line  whose 
angles  lie  in  the  support  verticals,  and  which,  for  the  case  of 
supports  on  a  level,  passes  through  the  corresponding  fixed 
points. 

It  follows  directly  that  the  moments  at  the  supports  are 
alternately  positive  and  negative,  and  increase  from  the 
end,  so  that  any  one  is  more  than  three  times  the  preceding. 
(See  Art.  111.) 

Since  now  this  polygon  has  alternately  angles  down  and  up, 
the  reactions  at  the  supports  must  be  alternately  positive  and 
negative.  From  the  corresponding  force  polygon  it  follows 
that  they  must  increase  from  the  end. 

The  shearing  forces  are,  therefore,  also  alternately  positive 
and  negative,  and  increase  from  the  end  on. 

95.  Loaded  Span. — Let  now  the  span  A  B  [PL  15,  Fig.  58] 
be  arbitrarily  loaded.  It  can  be  proved  here  also,  as  in  Art. 
92,  that  the  prolongation  of  the  sides  XT'  V  and  S  U,  as  also  of 
V"  U"  and  S  V,  intersect  in  the  limited  third  verticals. 

When  the  supports  are  in  a  straight  line,  then,  by  the  con- 
struction of  Art.  92,  the  fixed  points  I  and  K  are  the  intersec- 
tions of  S  V  and  S  U  with  A  B.  We  can,  therefore,  at  once 
assert,  that  the  sides  S  U  and  S  V  of  the  second  equilibrium 
polygon  pass  through  the  fixed  points  I  and-  K,  when  the  sup- 
ports are  on  a  level.  ^ 

For  known  position  of  the  fixed  points  and  for  given  load,  it 
is,  therefore,  easy  to  draw  the  second  polygon  by  drawing  ver- 
ticals H!  and  KKt  equal  to  the  corresponding  ordinates  of  the 
cross-lines.  S  U  and  S  V  pass,  then,  through  I  K^  and  K  It  re- 
spectively. Then,  by  Art.  89,  the  moments  at  the  supports  may 
be  determined. 

Since  A  I  <  -Z,  U  must  lie  to  the  right  of  I,  and  the  angle 
3 

AUS  is  concave  downwards.     Accordingly,  the  force  at  U, 
viz.,  -  M'  I,  acts  upwards.   The  same  holds  good  for  V.    Hence, 


144:  CONTINUOUS    GIRDER.  [CHAP.  IX 

the  moments  M'  M"  for  the  loaded  span  are  always  posi- 
tive*    If  we  draw  the  lines  M  N  and  U  V  cutting  the  middle 

vertical  inP  and  Q,  then,  for  b  =  \l,  P  O  =  1  (M'  +  M")  and 

o  2 

P  Q  =  2  (W  +  2ft")-    (See  Arts.  89, 90.)    Since  now  the  points 
U  and  V  must  lie  under  A  B, 

P  O  <  P  Q'  or  M'  -f  M''  <  3R'+  3B". 
If  the  supports  are  not  upon  a  level,  it  follows  from  this  Art, 
and  Art.  92,  that  the  intersections  of  S  U  and  S  V prolonged, 
with  the  prolongations  of  the  lines  A  A'  B  B',  joining  the  sup- 
ports of  two  adjacent  .spans,  lie  in  the  VERTICALS  THROUGH  THE 

FIXED  POINTS. 

96.    Two    successive    Loaded    Spans.— PL  15,  Fig.  59. 

1.  Here  also,  as  in  Art.  92,  we  can  prove  that  the  prolongar 
tions  ofSV  and  St  Uj.  intersect  in  the  limited  third  vertical. 

2.  Draw  through  B  a  line  which  intersects  S  V  and  Sj  U^  in 
I'  and  ri?  and  the  verticals  through  V,  W  and  Ut  in  V0,  W0 
and  U0. 

Then 

U0  Ut  :  V0  V  ; ;  U0  B  :  V0  B  ; :  Z0 :  Zx 

v0  v :  w0  w  : :  r  v0 :  r  w0. 

Hence  by  composition 


U  U  :  W  W 


:  r  v0  x  4  :  i'  w0  x  i± ; 

or  since  U0 1^  :  W0  W    :  U0 1\  :  W0 1\ 

UT    •  W  T  '  •     T'  V     v    7    •  T  "W     v    7 
01*  01     •  ~Q    A    Cfl  •  *      "  o     "*>     *!• 

If,  then,  the  point  I'  moves  in  a  vertical,  the  ratio  I'  V0  to 
I'  W0  does  not  change,  therefore  the  ratio  of  U0 1/  to  W0 I/  also 
remains  unchanged,  and  accordingly  I\  must  also  move  in  a 
vertical.  If  I'  coincides  with  I,  it  follows  from  the  construction 
of  Art.  93  that  the  point  I\  becomes  the  fixed  point  1^  Hence, 
the  intersections  T  and  I\  of  verticals  through  the  fixed  points 
I  and  Ii  with  the  sides  S  V  and  St  U\,  or  with  the  middle 
sides  of  the  two  polygons  adjacent  to  the  support,  lie  always 
in  a  straight  line  through  that  support,  for  any  heights  of 
supports. 

*  A  positive  moment  always  indicates  compression  in  lower  flange. 


CHAP.  IX.]       LOADED  AND  UNLOADED  SPANS.  14-5 

This  property  of  the  second  equilibrium  polygon  was  first 
made  known  by  Culmann. 

97.  Arbitrary  leading.  —  According  to  the  above  properties 
of  the  second  equilibrium  polygon,  the  general  course  of  pro- 
cedure for  any  given  case  of  loading  is  then  as  follows  [Fig.  60, 
PI.  16]  : 

1.  Construct  all  the  fixed  points  A  I2  13  ---  I^Ka,  etc.  (Art. 
93),  and  draw  verticals  through  them. 

2.  Construct  the  cross-lines  for  every  span,  Art.  88. 

3.  Make  A  C  equal  to  Ot  Qi  as  given  by  the  cross-lines,  and 
draw  a  line  through  C  and  Al  to  intersection  D2,  with  vertical 
through  I2.     Then  make  D2  C2  equal  to  O2  Q2,  and  draw  a  line 
through  C2  and  A2  to  D3,  and  so  on.     Precisely  the  same  con- 
struction holds  for  the  other  way  from  the  right  end.     Thus 
A4  E4  is  equal  to  R4  P4,  etc. 

4.  In  this  way  we  obtain  for  each  of  the  middle  sides  of  the 
second  equilibrium  polygon  two  points,  C  and  Fl5  C2  and  F2) 
etc.  ;  A  and  Et,  D2  and  E2,  and  so  on  ;  so  that  now  we  can  ac- 
tually draw  these  middle  sides. 

The  intersections  of  these  lines  with  the  support  verticals 
give,  according  to  Art.  88,  the  moments  at  the  supports.  For 
spans  whose  length  13  X  these  moments  are  given  directly;  for 
other  spans  the  ^  construction  of  Art.  74  must  be  applied.  The 
following  simple  construction  may  also  be  applied.  Let 
I  K  be  the  intersections  of  the  verticals  through  the  fixed 
points,  with  the  line  A  B  joining  the  supports  [Fig.  61,  PL  16]. 

Make    I  D'  =  I  D        *,   K  F'  =  KF     Y,    C'  D'  =  O  Q    - 


E'F'  =  RP  (^Y,  and  draw  C'  F'  and  E'  D'.     These  lines  cut 

\v  / 

the  support  verticals  in  M'  and  N',  so  that  A  M'  and  B  N'  are 
the  moments. 

By  the  construction  errors  accumulate  from  one  span  to  the 
next,  so  that  the  diagram  must  be  made  with  care.  We  have 
also  several  checks,  viz.  :  1.  The  intersection  of  the  middle  sides 
must  lie  in  the  vertical  through  the  intersection  of  the  cross- 
lines.  2.  The  prolongation  of  the  middle  sides  must  intersect 
in  the  limited  third  vertical.  3.  The  corresponding  intersec- 
tions of  the  middle  sides  with  the  third  verticals  must  lie  in  a 
straight  line  through  the  support. 
10 


146  CONTINUOUS    GIRDEK.  [CHAP.  IX. 

If  any  span  is  unloaded,  the  cross-lines  coincide.  The  above 
method  of  construction  holds  good  when  the  supports  are  not 
upon  a  level.  If  the  difference  of  height  of  the  supports  is 

represented  —  th  of  the  real  amount,  the  unit  for  the  moment 
m 

scale  is ,  as  is  easily  seen  by  reference  to  Arts.  81 

6  m  E  I 

and  88. 

The  above  method  of  construction  of  the  moments  at  sup- 
ports was  first  given  by  Culmann. 


CHAP.  X.]  SPECIAL   CASES   OF   LOADING.  147 


CHAPTEE    X. 


CONTINUOUS   GIRDEK  -  SPECIAL   CASES   OF   LOADING. 

98.  Total  uniform  Load.  —  If  a  span  is  loaded  with  a  uni- 
formly distributed  load  of  p  pounds  per  unit  of  length,  the 
simple  moment  area  is  a  parabolic  segment  whose  vertical  axis 
passes  through  the  centre  of  the  span.  [PI.  16,  Fig.  62.] 

1  -.-,.?••, 

The  ordinate  D  C"  is    p  P,  and  hence  the  area 


It  will  be  advantageous  here  to  take  p  X2  as  the  unit  of  the 
moment  scale  ;  and  therefore 


The  vertical  height  of  the  cross-lines  at  the  pole  distance  o 

from  the  middle  is  8ft  -.   If,  then,  we  take  b  =  -  X,  we  have 
X  6 


and  therefore 


2.  Moments. 

If  the  moments  A'  A"  and  B'  B"  are  known,  we  can  find 
the  end  tangents  of  the  first  equilibrium  polygon  by  drawing 
A"  B",  dropping  a  vertical  through  the  middle  D,  and  laying 

off  D  E  =  2  x  \p  P  =  \p  P  =  \p  X2  (|y.     The  lines  A"  E 

and  B"  E  are,  then,  these  end  tangents.     With  the  help  of 
these  we  may  easily  construct  the  parabola. 

3.  Shearing  force. 

If  we  draw  in  the  first  force  polygon,  lines  parallel  to  the 


148  CONTINUOUS    GIEDEE.  [dlAP.  X. 

end  tangents  and  the  closing  line  A'  B',  the  distances  on  the 
force  line  are  the  reactions  at  the  ends  of  the  span.  Instead 
of  this  we  may  lay  off  from  A  and  B,  A  G  and  B  H  equal  to 
the  first  pole  distance  #,  and  through  G  and  H  draw  parallels 
to  the  end  tangents,  intersecting  the  verticals  through  A  and  B 
in  A'  and  B^  We  thus  obtain  the  reactions  A  A1?  B  Bt.  The 
ordinates  to  Ax  B1  then  give  the  shearing  force  at  any  point. 

If  the  line  p  X2  representing  the  moment  units  is  equal  to  m, 
and  that  representing  the  force  units  p\  —  n^  then  the  first 

pole  distance  must  be  a  =  r-ra  X  =  —  X. 

2         m 


Accordingly,  it  is  now  easy,  from  the  general  construction 
given  in  Art.  97,  to  construct  the  shearing  force  and  moments 
for  uniform  or  dead  load  of  girder  in  any  case.  Let  us  pass 
on  to  an  example  illustrating  more  fully  the  above  principles. 

99.  Example.  —  As  an  example  of  the  application  of  the 
above  principles,  we  take  a  girder  of  four  spans,  as  given  in 
PL  17,  Fig.  63.  The  two  interior  spans  are  each  96  ft.,  the 
exterior  spans  80  ft.  each  ;  that  is,  l±  :  I  :  :  5  :  6.  Choose  any 
scale  of  length  convenient,  as,  for  instance,  50  ft.  to  an  inch, 
lay  off  the  spans  and  construct  first  the  fixed  points.  For  this 
purpose  we  draw  the  third  and  limited  third  verticals.  These 
last  are  easily  found  from  the  principle  already  deduced,  that 
they  must  divide  the  distance  between  the  third  verticals  into 
segments  inversely  as  the  corresponding  spans  [Art.  92].  Lay- 
ing off.  then,  from  the  third  vertical  in  the  first  span,  -J  I  to  the 
right,  or  from  the  third  vertical  in  the  second  span  -J  l±  to  the 
left,  we  have  the  first  limited  third  vertical.  The  same  at  the 
other  end  gives  .the  other.  For  the  centre  support,  of  course, 
the  limited  third,  since  the  adjacent  spans  are  equal,  passes 
through  the  support  itself.  We  can,  therefore,  now  construct 
i\\e  fixed  points  according  to  Art.  93. 

Let  the  load  per  unit  of  length  p  be  -J  ton  per  ft.  Then 
taking  X  [Art.  8§]  equal  to  Z,  we  have  n  =p\=pl  =  48  tons 
and  m=j>\2=pP=  4608  ft.  tons  [Art.  98  (3)].  It  remains 
to  assume  a  scale  of  force.  Let  this  be  20  tons  per  inch,  then 
our  moment  scale  is  20  x  50  =  1000  ft.  tons  per  inch.  The 
values  of  which  we  shall  need  to  make  use  are,  then,  to  scale 
^  =  1.6  inches,  X  =  I  —  1.92  inches, 


|)2  =  0.6944  inches,  (^  =  1.44  inches,  (|)*  =  0.4823  inches. 


CHAP.  X.]  SPECIAL    CASES    OF   LOADING.  149 

These  values  are  repeated  upon  the  PL  for  convenience  of 
reference.  Also,  p  I  =  48  tons  —  2.4  in.  =  n,  p  fi  =  4608  ft. 
tons  =  4.608  in.  =  m.  For  the  first  pole  distance  [Art.  98  (3)] 

n  2.4  25 

we  have  a  =  —  X  =  .  -A:>  I  —  -^  I  =  1  in.      Second  pole  dis- 
m         4.608         48  r 

tance  [Art.  98]  b  =  ^  X  -  0.32  in. 

According  to  Art.  98,  we  have  now,  for  the  cross  lines, 

:?X»/        and  O'  P'  =  Q'  P'  = 


Laying  off  these  distances  under  the  supports,  we  have  thus  the 
cross-lines. 

We  have  next  to  construct  the  second  equilibrium  polygon. 
This,  by  the  aid  of  the  cross-lines  and  fixed  points  already  con- 
structed, we  can  easily  do,  as  detailed  in  Art.  97  (3).  Then 
the  moments  at  the  supports  are  given  directly  to  moment 
scale  in  the  interior  spans,  or  we  can  find  them  from  the  end 

spans  by  laying  off  £  j  I  [Art.  89]. 
h 

Finally,  the  moments  thus  found  and  laid  off  at  the  sup- 
ports, we  can  construct  the  moment  curve  by  making  D'  E'  — 

/  \2  fl\2 

r     and  DE  =  i^X2-     [Art.  98  (2)],  and  thus  draw- 


ing the  end  tangents  and  corresponding  parabolas. 

According  to  Art.  98  (3),  we  can  then  find  the  shearing 

A-I  4 

forces  by  laying  off  a  =  —  A  and  drawing  parallels  to  the  end 

tangents  to  intersection  with  verticals  through  supports,  as 
shown  in  Fig. 

We  thus  have  both  moments  and  shearing  forces  for  uniform 
load.  By  careful  attention  to  the  above,  the  reader  will  have 
no  difficulty  in  solving  any  case.  We  recommend  him  earn- 
estly to  perform  the  entire  construction  for  himself,  referring 
to  the  proper  Arts,  at  every  step.  [For  convenience  of  size, 
we  have  not  observed  our  scales  strictly  in  the  Figs.  TJie 
reader  should  therefore  ?wt  attempt  to  check  results  with  the 
dividers.'} 

1OO.  Partial  uniform  Load.  —  1.  When  the  girder  is  only 
partially  loaded,  as,  for  instance,  a  certain  portion  of  the  span 
id  I  from  B,  the  simple  moment  area  consists  of  a  triangle  ABC 


150  CONTINUOUS    GIEDEK.  [CHAP  X. 

and  a  parabolic  segment  C  E  B.  [PI.  16,  Fig.  64.]  If  in  the 
first  force  polygon  B'  D'  is  the  total  load  upon  the  span,  the 
line  O  A'  parallel  to  the  end  tangent  A  G  divides  B'  D'  in  the 
same  ratio  as  the  end  of  the  load  divides  the  span,  or  denoting 
the  length  B0  C0  by  /3  1. 

B'  A'  :B'D  '  :;/3l:l. 

The  intersection  G  of  the  end  tangents  lies  in  the  vertical 
G  H,  which  halves  B  F.  Since  the  triangle  B  C  T  is  similar  to 
O  A!  B',  we  have 

BT:B'  A'  :;££:#;     • 

a 

or  since  B'  D'  =p  I,     B'  A'  =p  j3  =  <pl. 


It  is  therefore  easy  to  construct  B  T  as  in  Fig.  64,  where 

/Q 

B!  D!  —p  I,  A!  Bt  =  B'  A'  =  p  I  7,  and  pole  distance  =  i  £  ; 

then  B2  D2  =  B  T. 

If  B  T  is  thus  found,  we  can  easily,  when  A  and  B  are  given, 
construct  the  end  tangents,  and  then  construct  the  first  equili- 
brium curve  itself. 

2.  Make  G  K  —  £  GI,  then  the  triangle  CKB  is  equal  to 
the  parabolic  area  C  E  B.     If,  then,  through  K  we  draw  a 
parallel  to  C  B,  intersecting  C  G  in  L,  and  through  L  the  ver- 
tical LM,  the  triangle  ALB  is  equal  to  the  entire  simple 
moment  area.     This  last  is  therefore  proportional  to  L  M,  or 
2ft  I  =  $  I  x  L.  M  ;   hence  %tt  =  J  L  M.     It  is  easily  proved 
also  that  F  M  =  i  F  B.     Thus,  as  GK  is  £  of  GI,  GL  is 
J-  of  G  C  ;  hence  M  H  is  $  of  F  H,  and  therefore  F  M  is  f  of 
F  H,  or  |  of  J-  F  B  =  ^  F  B.     L  M  can  therefore  be  easily 
drawn. 

3.  Let  N  be  the  middle  of  A  B.     Make  N  O  equal  to  $  F  N. 
Then  the  centre  of  gravity  of  the  triangle  A  C  B  is  in  the  ver- 
tical through  O,  while  the  centre  of  gravity  of  the  parabola  is 
in  H  G.     If  through  L  we  draw  a  parallel  to  AB,  intersecting 
the  vertical  through  C  in  P,  the  two  areas  ^are  to  each  other 
as  FC  to  C^P.     If  in  the  vertical  through  6  we  make  OQ  = 
J  CP,  then,  since  IH  =  J  CF,  we  have 

IH:OQ::FC:CP. 


CHAP.  X.]  SPECIAL   CASES   OF"  LOADING.  151 

The  intersection  R  of  the  line  Q  I  with  A  B  lies,  then,  in  the 
vertical  through  the  centre  of  gravity  of  the  simple  moment 
area. 

Thus  the  construction  of  the  cross-lines  is  now  easy. 

4.  It  is  most  convenient  in  the  application  of  the  above  to 
construct  or  calculate  the  distances  of  the  cross-lines  under  the 
supports  once  for  all,  for  load  over  various  parts  of  the  span. 
The  necessary  formulae  can  be  directly  deduced.  Thus  the 


Triangle  B  AT  =  BT  x      l  =  ~       X 

\$  1=^  x 


Triangle        BC  T  =  BT  x      $  1=          x      01. 


Triangle       BLT  =  BTX/3Z  =xjS  I 

6  A  (t,      6 

The  entire  area  is  equal  to  the  triangle  ALB. 
But  A  L  B  =  B  A  T-B  L  T  =  2&  (I  Z-l  ft 

2t  d    \^          O 

The  triangle  A  C  B  =  B  A  T—  B  C  T  ;  hence 


The  parabolic  area  is  equal  to  the  entire  area  minus  A  C  B, 

7}  BP      1 

or  parabolic  area   =  TT—  *•#  P  I- 

2i  Cb        O 

F  is  distant  from  B  T  by  a  distance  ft  I,  N  by  a  distance  = 

\l,  NO  is  \  of  N  F  ;  hence  O  is  distant  from  B  T 
2i  o 


Therefore,  the  moment  of  the  triangular  area,  with  reference 
to  the  right  support  B,  is 


The  moment  of  the  parabolic  area  is 


The  total  moment,  with  reference  to  the  right  support,  is, 
therefore, 


152  CONTINUOUS    GIRDER.  [CHAP.  X. 

In  a  similar  manner  we  find  for  the  left  support 


When  ft  =  1  the  span  is  completely  covered,  and  we  have, 
then,  right  and  left 


24  a 

If  we  compare  this  value  with  those  for  partial  loading,  we 
see  that  they  differ  only  by  certain  coefficients,  and  that  these 
coefficients  depend  only  upon  the  length  of  the  loaded  portion. 
If,  then,  we  have  the  distance  between  the  cross-lines  for  total 
load,  we  have  only  to  multiply  by  certain  factors  to  obtain  the 
distances  for  partial  loading.  For  uniform  or  total  load  over 

1          /  A4 
the  whole  span,  this  distance  is  given  by  -p\2  I  - )    (Art.   98). 

4:  \  A// 

If  we  divide  this  distance  in  certain  proportions  we  have  at 
once  the  distances  for  partial  loading.  These  proportions  are 
given  by  (2— 0*)  $  for  the  right  support,  and  (2-/3)2  /3  for  the 
left,  under  the  supposition  that  .the  load  comes  on  from  the 
right.  The  reverse  is  the  case  for  load  coming  on  from  left. 

~|       ~|       Q 

If  we  take  /3  =  -,-,-  of  the  span,  we  can  calculate  these  pro- 
4  2i  4 

portions  once  for  all.     We  thus  have  the  following  table : 

Support  under  load  Support  for  wwloaded  end 

(2-02)  /5».  (2-p)»  p. 

-  span  loaded — —  =  0.1211 . . . 


i  span  loaded — |—  =  0.4375 ....  —  =  0.5625. 


3  207  225 

-  span  loaded —  =  0.8086.  ..:-&&**  0.8789. 


The  division  of  the  distance  between  the  cross-lines  for  uni- 
form load  over  the  whole  span  into  these  proportions  is  easily 
accomplished  graphically.  Thus,  from  the  end  of  the  line  to 
be  divided,  draw  a  line  in  any  direction,  and  lay  off  upon  it  the 
six  numbers  above,  to  any  convenient  scale.  Join  the  end  of 
the  last  division  with  the  end  of  the  line  to  be  divided,  and 
then  draw  parallels  through  the  other  points. 

It  is  important  to  observe  some  definite  system  of  numera- 
tion, otherwise,  especially  in  the  first  attempt  at  construction, 
confusion  is  apt  to  arise. 


CHAP.  X.]  SPECIAL    CASES    OF   LOADING.  153 

We  can  thus  find  the  cross-lines  for  any  position  of  the  load, 
and  for  each  position  can,  if  we  wish,  draw  the  equilibrium 
polygon  and  determine  the  moments  according  to  the  general 
method  of  Art.  97. 

1O1.  Concentrated  Load.  —  The  simple  moment  area  is  in 

this  case  a  triangle   [PL  16,  Fig.  65]  whose  area  is  -  I  h,  h  be- 

2 

ing  the  height  C  D.     Therefore 


If  from  the  centre  of  the  span  E  we  lay  off  E  F  =  -  E  D,  D 

o 

being  the  point  of  application,  a  vertical  through  F  passes 
through  the  centre  of  gravity. 

As  the  height  C  D  is  proportional  to  9Cft,  we  may  take  C  D  as 
second  force  polygon.  Since  h  =.  2  £0?,  the  distance  of  the  pole 

N  must  be  2  x  \  I  =  i  Z,  when  X  =  I  (Art.  89).     Draw  N  P 
6         3 

parallel  to  A  B,  then  is  N  P  =  -  A  B. 

3 

Parallel  to  N  C  and  N  D  we  may  draw  the  cross-lines.  A 
simple  construction  may  be  given  for  them  when  they  are  made 
to  pass  through  A  and  B.  Let  A  M  and  B  L  be  the  cross-lines. 
Then  the  triangle  S  B  M  is  similar  to  N  C  D,  and 

B  M  :  C  D  ;  ;  B  F  :  N  P.      But 


and  N  P  =  £  A  B;  therefore, 

B  M  :  C  D  ;  ;  (2  A  B-A  D)  :  A  B. 
*Make  D  G  =  A  B,  then  B  G  =  2  A  B-A  D. 

The  point  M  is  accordingly  found  by  drawing  a  straight  line 
through  G  and  C,  D  G  being  equal  to  A  B.  In  the  same  way 
make  D  H  =  A  B,  and  draw  a  line  through  H  and  C.  We 
thus  obtain  L. 

The  prolongations  ofM.C  and  I*  C,  therefore,  intersect  A  B 
prolonged  in  the  points  G  and  H,  distant  from  D  ~by  A  B. 

If,  then,  we  have  to  investigate  a  concentrated  load  in  various 
positions,  we  draw  the  first  equilibrium  polygon  C  X  and  C  Y 
[PI.  16,  Fig.  66],  and  lay  off  in  the  same  the  closing  lines  (for 


154  CONTINUOUS    GIRDER.  [CHAP.  X. 

the  simple  moment  area)  for  the  different  positions  of  the  load. 
The  distances  cut  off  on  the  vertical  through  C  by  these  lines 
give,  then,  the  various  values  of  h  or  2  %R.  If  we  draw  from. 

these  intersections  lines  to  the  pole  N,  at  the  distance  -  I  from 

3 

D  C,  the  cross-lines  are  parallel  to  these  lines.  It  will  be  best 
to  keep  for  each  pair  the  common  line  P  Q  parallel  to  N  C. 
"When  the  point  of  application  of  the  load  divides  the  span  into 
two  equal  parts,  the  point  of  intersection  of  the  cross-lines  di- 
vides the  middle  third  of  P  Q  into  equal  parts. 

The  centre  of  gravity  of  the  simple  moment  area  cannot  pass 
beyond  the  middle  third  of  the  span.  Since  any  load  can  be 
considered  as  made  up  of  a  number  of  concentrated  loads,  it 
follows  generally  that  far  any  method  of  loading  the  centre  of 
gravity  of  the  simple  moment  area  lies  between  the  third  verti- 


CHAP.  XI.]  MAXIMUM   STRAINS.  155 


CHAPTEE  XI. 

METHODS   OF   LOADIN0   CAUSING   MAXIMUM   STRAINS. 

1O2.  Maximum  Shearing  Force—  -Uniformly  distributed 
Jloviiii?  Load.  —  Suppose,  first,  the  span  in  question  loaded 
with  a  concentrated  weight.  The  simple  moment  area  is 
A'  C'  B'.  [PL  18,  Fig.  67.] 

In  the  force  polygon  let  O  A1?  O  Bx  and  O  Ot  be  respectively 
parallel  to  C'  A',  C'B'  and  A7  B7.  Then  ^  Aj  and  C^  are 
the  reactions  at  A  and  B.  Since,  according  to  Art.  80,  the  mo- 
ments A  A7  and  B  B'  are  always  positive,  and  the  middle  sides 
A'  S,  B7  S  pass  through  the  fixed  points  1  and  K,  it  follows 
from  the  construction  of  the  preceding  Art.  that  the  intersec- 
tions O  and  P  of  the  sides  A'  C'  and  B'  V  of  the  first  equili- 
brium polygon  with  the  closing  line  A  B  must  always  lie  within 
A  I  and  B  K.  That  is,  the  points  of  inflection  O  and  K  are  al- 
ways between  the  fixed  points  and  the  ends.  Therefore  A',  B' 
and  the  point  C'  must  lie  on  opposite  sides  of  the  closing  line 
A  B,  and  consequently  Cj.  in  the  force  polygon  must  lie  between 
A!  and  B^ 

Accordingly  the  shear  At  C^  at  A  is  positive^  and  the  shear 
G!  B1  at  B  is  negative. 

Let  the  distance  of  the  load  from  the  left  support  be  /3,  from 
the  right  support  &.     The  load  itself  is  P,  and  the  moment 
A  A'  at  the  left  support  M',  B  B'  at  the  right  M".     Kequired, 
the  shearing  force  S  at  a  point  distant  x  from  the  left  support. 
The  partial  reaction  at  the  left  is  R'.     Then 
R'  Z  =  M'  -  M"  +P&, 
M'^M" 

~~  ~ 


M'  and  M"  are  always  positive.  If,  therefore,  M'  >  M",  R' 
is  positive  ;  if  M'<  M",  then  M"  -  M'<  M"  +  M7,  or,  since 
by  Arts.  75  and  80,  M'  +  M"<  2R'  +  3JT,  M"  -  M^aft'+aR". 
Now  it  can  be  easily  proved  analytically  that  for  a  girder  hori- 


zontally fixed,*  mf  =    ~  and  2R"=  3  hence  3B'  +  3R" 


Supplement  to  Chap.  VII.,  Art.  18. 


156  METHODS  OF  LOADING  [CHAP.  XI. 

since  £  +  &  =  J.      Therefore   M"  -  M' 


Since,  however,  /8  <  Z,  we  have  also  M"—  M'  <  P  $_.  Hence,  if 
M'  is  <  M",  we  have  also  R'  positive.  R'  is  therefore  always 
positive  whatever  may  be  the  position  of  the  load.  In  the  same 
way  it  may  be  shown  that  R"  is  always  negative. 

If  now  the  load  is  to  the  right  of  the  point  distant  x  from 
the  left  support,  then  for  this  point  the  shearing  force  S'  —  R', 
and  is  therefore  positive.  If  the  load  is  to  the  left  of  this 
point,  the  shearing  force  S  =  R'  —  P  —  R",  and  is  therefore 
negative.  S  for  any  point  is  therefore  positive  or  negative, 
according  as  the  load  lies  right  or  left  of  this  point.  Hence 
for  a  uniform  load  we  deduce  directly  — 

The  shearing  force  at  any  point  is  a  positive  or  negative 
maximum  when  the  load  extends  from  this  point  to  the  right 
or  left  support  respectively. 

The  same  principle  holds  good  for  the  simple  girder. 

2.  Thus  far  we  have  considered  the  load  in  the  span  itself. 
Suppose  now  the  load  is  in  some  other  span,  and  the  span  in 
question  is  unloaded,  then 

M'  -  M"         „      M"-M'" 

R  =-  -IT     R  =      r  • 

As  we  pass  away  from  the  loaded  span  the  moments  at  the 
supports  are  alternately  positive  and  negative,  and  each  is 
greater  than  the  one  following  (Art.  94).  Since  the  moments 
M'  and  M"  are  alternately  positive  and  negative,  R'  will  have 
the  same  sign  as  M',  and  R"  as  M",  and  generally  R^  as  Mm. 

Adopting,  then,  the.  notation  shown  in  PL  18,  Fig.  68,  we 
have  for  the  span  lm_± 

.t  -  Mm 


where  Rra  has  the  same  sign  as  Mm. 
In  the  same  way 


and  therefore 

1    _ 


CIIA1'.  XI.]  CAUSING   MAXIMUM    STRAINS.  157 

M 

But      rc+    is  negative  and  greater  than  2  (Art.  94-).     There- 
fore in  the  preceding  expression  the  numerator  is  positive  and 

ivi  1 

>  3.     Further,  -«^— L  is  negative  and  less  than  -,  hence  the  de- 

1V1m 

•  3 

nominator  of  the  above  expression  is  negative  and  <  -.    There- 

R  I 

fore    R+1  is  negative  and  >  2  -y1^,  that  is,  the  shearing  forces 

at  the  supports  are  alternately  positive  and  negative,  and  in- 
crease (when  2  lm_±  is  not  less  than  Zm)  towards  the  loaded  span. 
We  have  then 


For  any  span,  then,  the  shear  at  the  left  support  R'  will  be 
positive  when  the  left  adjacent  span  is  loaded,  the  right  adja- 
cent span  unloaded,  and  all  the  other  spans  each  way  alternately 
loaded.  The  shear  R'  will  be  negative  when  the  remaining 
spans  are  loaded.  Hence  : 

The  shearing  force  is  a  maximum  (positive)  at  any  point 
when  the  load  extends  from  this  point  to  the  right  support,  and 
the  other  spans  are  alternately  loaded,  the  adjacent  span  to  the 
right  ~being  unloaded,  that  to  the  left,  loaded.  The  negative 
maximum,  on  the  contrary,  occurs  when  the  load  extends  from, 
the  point  to  the  left  support,  when  the  right  adjacent  span  is 
loaded  and  the  left  unloaded ;  the  other  spans  alternately 
loaded. 

PL  18,  Fig.  69,  gives  these  two  cases. 

In  practice  such  a  loading  can  never  occur.  If  we  suppose 
the  rolling  load  divided  into  two  portions  only,  the  above  rule 
reads  as  follows : 

The  shearing  force  at  any  point  will  be  a  positive  maximum 
when  the  load  reaches  from  the  right  support  to  this  point,  and 
when  the  left  adjacent  span  is  covered.  The  negative  maximum 
occurs  when  the  load  reaches  from  left  support  to  the  point, 
and  the  right  adjacent  span  is  covered,. 

1O3.  Maximum  Moments. — 1st.  Loaded  Span.  Let  a  weight 
act  at  the  point  D,  PL  18,  Fig.  67.  Then  A  U  S  V  B  is  the 
second.  A'  C'  B'  the  first  equilibrium  polygon,  and  A  A',  B  B' 


158  METHODS    OF   LOADING-  [CHAP.  XI. 

are  the  moments  at  the  supports.  We  have  already  seen  that 
the  inflection  points  O  and  P  (for  which  the  moment  is  zero) 
lie  outside  of  the  fixed  points.  We  can  therefore  assert  that 
WITHIN  the  fixed  points  the  'moments  are  negative  wherever  the 
weight  may  Replaced.  From  the  Fig.  we  see  at  once  that  the 
inflection  points  O  and  P  move  to  the  right  or  left  as  the  weight 
moves  to  the  right  or  left.  Accordingly  when  for  the  weight 
at  D  the  moment  at  O  is  zero,  the  moment  at  this  point  will  be 
positive  when  the  load  moves  to  the  right  of  D,  negative  when 
it  moves  to  the  left  of  D. 

Hence  for  the  maximum  moment  we  have  at  once  the  follow- 
ing principle : 

For  any  point  O  outside  the  fixed  points  the  moment  will  ~be 
a  positive  or  negative  maximum  when  the  load  reaches  from 
the  point  D,  where  a  load  must  Replaced  to  cause  the  moment 
at  O  to  ~be  zero,  to  the  right  or  left  support  respectively.  For 
the  negative  maximum,  therefore,  the  load  reaches  from  A  to  D; 
for  the  positive,  from  D  to  B. 

If  the  point  O  is  given,  it  is  indeed  possible  to  determine  by 
construction  the  point  D  to  which  the  load  must  reach.  It  is, 
however,  simpler  to  assume  D  and  then  construct  O. 

If  we  choose  for  the  different  positions  of  D  an  arbitrary 
length  for  C'  D'  (Fig.  67),  so  that  the  point  C'  falls  in  a  parallel 
Q  R  to  A'  B'  (Fig.  70),  and,  moreover,  take  D  at  equal  intervals, 
then  the  points  L  and  M  will  be  at  equal  distances  (Fig.  67), 
and  hence  the  points  I  and  K  (Fig.  70),  in  which  the  verticals 
through  the  fixed  points  are  intersected  by  the  lines  A'  M  and 
B'  L  (Fig.  67),  will  be  at  equal  distances.  We  have,  then,  the 
following  simple  construction  [Fig.  70,  PL  18] : 

Between  the  verticals  through  the  supports  draw  two  parallel 
lines  A  B  and  Q  R  at  any  convenient  distance  apart,  and  divide 
Q  R  into  a  number  of  equal  parts ;  four  or  five  are  sufficient. 
Draw  lines  from  A  to  R  and  the  middle  S  intersecting  the  ver- 
tical through  the  fixed  point  I  in  Ix  and  I2.  In  the  same  way  find 
K!  and  K2.  Divide  It  I2  and  Kt  K2  into  the  same  number  of 
equal  parts  as  Q  R  has  been  divided  into,  and  join  these  points 
in  reverse  order  by  lines.  The  intersection  of  these  lines  with 
the  lines  drawn  from  A  and  B  to  the  points  upon  Q  R  give  the 
points  O,  for  which  the  moment  is  a  maximum  when  the  load 
is  limited  by  the  corresponding  point  upon  Q  R. 

This  construction  was  first  given  by  Mohr. 


CHAP.  XI.]  CAUSING  MAXIMUM   STRAINS.  159 

IO 1.  Determination  of  the  maximum  Shearing  Forces. 

— According  to  the  general  method  of  construction  given  in 
Art.  97,  we  can  now  determine  by  reference  to  Arts.  98  and  99, 
which  treat  of  total  and  partial  distributed  loading,  the  shear- 
ing forces  corresponding  to  the  methods  of  loading  which  cause 
maximum  strains. 

As  a  review  of  the  preceding  principles,  we  take  the  same 
example  as  before,  as  given  in  PL  19,  Fig.  71.  Here  again  the 
reader  should  construct  the  Figs,  for  himself.  The  scales  are 
as  before,  Art.  99. 

Fig.  a  shows  the  method  of  loading  for  positive  maximum 
shear  in  first  span ;  and  the  second  Fig.  below,  the  same  for 
the  second  span.  [Art.  102.]  , 

We  first  find,  precisely  as  in  Fig.  63,  the  shearing  forces  in 
the  third  and  fourth  spans  for  the  total  loads  over  those  spans, 
and  lay  off  the  shear  thus  obtained  in  the  first  and  second  spans, 
as  indicated  by  the  broken  lines  in  Fig.  b  in  those  spans.  Thus 
having  first  found  the  fixed  points,  which  we  may  here  take 
directly  from  Fig.  63,  we  construct  as  in  that  Fig.  also  the  cross- 
lines  for  total  load  in  third  and  fourth  spans.  Thus  laying  off 
54  equal  to  Oj.  P1?  and  drawing  a  line  through  4  and  support  to 
intersection  with  vertical  through  fixed  point  in  second  span 
[Fig.  60],  we  determine  D',  and  then  from  the  cross-lines  find 
at  once  D".  In  like,  manner,  supposing  for  the  moment  the 
load  on  the  other  two  spans,  we  have  e  a,  a  D,  D  a  and  a  F', 
and  then  at  once  F''.  F"  D'  cuts  off  then  the  moment  at  the 
right  support,  and  joining  1,  2  and  5,  we  find,  according  to  Art. 
98  (2),  precisely  as  in  Fig.  63,  the  end  tangents ;  and  then  from 
these,  with  the  first  pole  distance  a,  find  the  shear.  This  is 
given  by  the  broken  lines  in  third  and  fourth  spans.  Lay  off 
these  lines  in  first  and  second  spans,  remembering,  since  the 
shears  at  the  supports  alternate,  that  the  positive  shear  at  left 
of  fourth  span  must  be  laid  off  as  negative  (down)  at  right  end 
of  first,  etc.  [See  Fig.  63.] 

Now  for  the  positive  shears  in  first  and  second  spans  for  the 
different  methods  of  loading,  we  have  only  to  determine  the 
direction  of  the  tangents  through  end  of  load  [Fig.  64,  Art. 
100].  The  lengths  of  the  segments  cut  off  upon  the  verticals 
through  the  supports  by  these  tangents  are  given  for  first  and 
second  spans  by  Figs,  /and  e,  for  each  position  of  load.  [Art. 
100  (1).] 


160  METHODS  OF  LOADING  [dlAP.  XI. 

We  have  first,  then,  to  construct  the  cross-lines  [Art.  100  (4)] 
as  shown  in  Fig.  d. 

.  Take  first  the  second  span.  For  this  span,  as  shown  in  Fig. 
dj  the  first  span  is  fully  loaded,  as  also  the  last.  Make,  there- 
fore, e  a  =  Oi  P1?  draw  a  D,  and  we  thus  find  the  point  D,  com- 
mon to  all  the  middle  sides  of  the  second  equilibrium  polygon 
for  different  positions  of  load.  So  also  make  on  right  54  — 
Ot  P!  and  draw  4  D',  and  then,  since  third  span  is  empty,  D'  F, 
and  we  thus  have  P.  Now  from  D  and  F  lay  off  the  cross- 
line  distances,  as  "D  a,  D  5,  D  <?,  equal  to  e  a,  e  b,  e  <?,  etc.  Draw 
lines  through  these  points  and  D  and  F  respectively,  and  note 
their  points  of  intersection  abed  with  the  verticals  through 
the  supports.  [NOTE.  Be  careful  to  preserve  an  orderly  nota- 
tion.] These  points  give  the  moments  for  each  position  of 
load  in  second  span.  Take  the  length  a  a  from  Fig.  e  and  lay 
it  off  from  a  on  the  right  support  vertical,  and  join  the  end 
with  a  on  left.  This  is  the  tangent  for  full  load  in  second 
span.  A  parallel  to  it  at  distance  a  from  left  gives  the  shear 
in  Fig.  b.  Then  lay  off  b  b  taken  from  Fig.  e  on  right,  and  join 
with  b  on  left.  This  is  tangent  for  load  over  three-fourths 
second  span  from  right.  A  parallel  to  it  at  distance  a  from 
foot  of  perpendicular  one-fourth  of  span  from  left  cuts  off 
shear  for  this  position  of  load.  So  for  tangents  c  c,  d  d.  We 
thus  obtain  the  curve  for  positive  shears  in  second  span.  The 
negative  shears  are  obtained  by  subtracting  these  from  the 
shear  already  found  for  full  load.  We  thus  have  the  lower 
curve,  and  the  shear  diagram  for  second  span  is  complete. 

For  first  span,  only  the  third  is  loaded.  We  lay  off,  then, 
78  equal  to  the  distance  between  cross-lines  corresponding, 
draw  8  F1?  and  thus  find  Ft.  Lay  off  now  at  left  end  ed,  ec, 
e  by  draw  lines  from  these  points  through  second  support,  and 
note  intersections  with  vertical  through  D.  Through  each  of 
these  intersections  draw  a  line  through  F1?  and  produce  to 
intersections  abed  with  vertical  through  second  support.  It 
is  from  these  last  points  that  the  distances  a  a,  bb,  etc.,  taken 
from  Fig.  f  must  be  laid  off  respectively  in  order  to  find  the 
tangents  e  a,  e  b,  e  c,  e  d,  e  e. 

Parallels  to  these  tangents  above  in  Fig.  b  give,  as  before, 
the  positive  shear  for  each  position  of  load.  The  negative 
shear  is,  as  before,  found  by  subtracting  the  positive  from  total 
load  shear.  Thus  shear  diagram  for  first  span  is  complete. 


CHAP.  XI.]  CAUSING   MAXIMUM    STRAINS.  161 

Of  course,  the  same  circumstances  can  hold  good  for  third 
and  fourth  spans  as  for  first  and  second,  except  that  positive 
shears  on  one  side  of  centre  support  are  the  corresponding 
negative  shears  upon  the  other,  and  vice  versa. 

Following  carefully  the  above  with  the  aid  of  the  Fig.,  the 
reader  cannot  fail  to  grasp  the  method.  An  independent  con- 
struction for  a  similar  case  will  make  both  principles  and  de- 
tails familiar.  Once  thoroughly  understood,  the  method  is 
rapid,  accurate,  simple,  and  of  general  application. 

1O5.  ]>etcr  111  illation  of  the  Maximum  Moments. — In 
like  manner,  it  is  easy,  according  to  the  general  construction 
given  in  Art.  97,  and  referring  to  Arts.  98  and  99,  to  determine 
the  maximum  moments.  In  Fig.  72,  PL  20,  we  have  the  same 
example  as  before,  concerning  which  we  have  but  little  addi- 
tional to  remark.  Fig.  64,  Art.  100,  shows  that  the  end  tan- 
gents give  the  moments  within  the  unloaded  portion  of  the 
girder.  These  tangents  are  constructed  precisely  as  before  in 
the  several  spans,  except  it  will  be  noticed  that  in  the  first 
span  we  have  made  use  of  the  construction  given  in  Art.  97, 
Fig.  61.  Thus  the  point  F'  is  determined  so  that  K  F'  = 

(A2 
j-] ,  and  thus  the  moments  are  measured  directly  at  the 

end  vertical.     Also  upon  the  left  support  vertical  we  have  laid 
off    the  distances  between  the  cross-lines  in  the  second  span 

(7  \2 
7/  ' 

The  only  thing  new  in  the  PL  is  Fig.  c,  which,  as  we  have 
seen  in  Art.  103,  Fig.  70,  gives  the  points  at  which  the  positive 
moment  is  a  maximum  for  each  position  of  load..  The  positive 
moments  can  then  be  taken  directty  off  upon  the  verticals 
through  these  points,  and  are  limited  by  the  horizontal  through 
the  supports  and  the  tangents,  as  above. 

Thus,  at  second  support  the  vertical  distance  to  a,  for  first 
span,  gives  the  moment  at  the  support.  La}r  it  off  in  Fig.  a 
from  the  support  line.  For  a  load  over  f  span,  wre  see  at  once 
the  point  for  which  the  positive  moment  is  a  maximum  from 
Fig.  c.  Follow  up  the  vertical  through  this  point.  The  dis- 
tance on  this  vertical  in  Fig.  J,  between  the  support  line  and 
tangent  b  b^  gives  the  moment  to  be  laid  off  in  Fig.  a  upon 
this  vertical.  So,  for  load  over  £  span,  we  have  next  vertical 

and  tangent  cc^  and  so  on.     We  thus  obtain  the  curve  for 
11 


162  METHODS  OF  LOADING  [CHAP.  XI. 

• 

positive  moments  at  the  right  end  of  first  span,  aticde.  From 
e  draw  a  line  to  left  support.  At  the  other  supports,  in  like 
manner,  we  determine  the 'positive  moments,  and  join  the 
points  ee  in  second  and  third  spans,  and  e  and  right  support 
in  fourth. 

We  have  already  seen  (Art.  103)  that  within  the  fixed  points 
the  moments  are  negative  wherever  a  load  may  be  placed. 
The  maximum,  therefore,  occurs  for  full  load.  We  have,  there- 
fore, found  for  3d  and  4th  spans  the  parabola  for  full  load, 
precisely  as  in  Fig.  63.  These  parabolas  are  given  in  broken 
lines  in  the  Fig.  a.  By  subtraction  of  the  positive  moments 
outside  the  fixed  points  from  the  positive  moments  at  the  same 
points  for  total  load  given  by  these  parabolas,  we  obtain  directly 
the  lower  curves  as  far  as  the  points  e  in  each  span. 

The  second  parabola  (partly  full,  partly  broken)  is  all  that  is 
needed  to  complete  our  Fig.  To  obtain  this  we  have  simply, 
in  2d  and  3d  spans,  to  make  the  vertical  through  centre  of 
line  e  e  equal  to  its  length  already  laid  off  for  total  load,  viz. : 

Z\2 

-J,  Art.  98  (2),  produce  e  e  to  intersections  with  sup- 
port verticals,  and  join  these  intersections  with  the  extremities 
of  the  first  verticals  above.  We  can  then  construct  the  para- 
bola, which  completes  our  diagram,  and  gives  us  Fig.  a. 

1O6.  Practical  simplifications. — In  practice,  the  construc- 
tions given  in  Figs.  63,  71  and  72,  admit  of  many  simplifica- 
tions, which,  in  order  to  avoid  confusion  at  first,  have  been  di& 
regarded.  The  whole  solution,  given  for  the  sake  of  clearness 
in  three  separate  Plates,  can  be  performed  upon  a  single  sheet, 
since  the  Fig.  for  the  second  equilibrium  polygon  in  Figs.  63, 
71  and  72  may  be  combined  in  one.  Indeed,  the  lines  neces- 
sary for  the  construction  of  the  maximum  shearing  forces  can 
be  applied  directly  to  the  determination  of  the  maximum 
moments.  It  is  therefore  unnecessary  to  divide  the  construc- 
tion into  separate  sheets. 

2.  The  cross-lines  in  the  end  spans  can  be  omitted,  since 
all  that  is  required  are  the  distances  to  be  laid  off  upon  the 
end  verticals,  and  these  when  found  can  be  laid  off  at  once. 

3.  We  can  apply  the  second  equilibrium  polygon  directly 
in  order  to  find  the  moments  for  the  dead  and  moving  load. 
Thus  the  transferring  of  ordinates  from  one  Fig.  to  another  is 
avoided. 


CHAP.  XI.]  CAUSING   MAXIMUM   STRAINS.  163 

4.  It  is  evidently  unnecessary  to  actually  draw  all  the  vari- 
ous lines.     We  need  only  to  mark  the  different  points  of  inter- 
section. 

5.  The  construction  for  dead  and  live  loads  can  be  per- 
formed at  once,  thus  avoiding  the  necessity  of  a  subsequent 
addition. 

1O7.  Approximate  Practical  Constructions  —  If  the  suc- 
cessive steps  of  the  preceding  development  are  carefully  fol- 
lowed, the  method  will  be  found  simple  and  easy  of  appli- 
cation. Indeed,  the  complete  and  accurate  solution  of  the 
difficult  problem  of  the  continuous  girder  by  a  method  purely 
graphical,  is  the  most  important  extension  of  the  system  since 
the  date  of  Culmann's  treatise,  and  well  illustrates  the  power 
and  practical  value  of  the  Graphical  Method. 

Humber  gives  the  following  constructions,  "  which  may  be 
relied  upon  for  safety  without  extravagance."  *  As  rapid 
means  of  obtaining  approximate  results,  they  may  not  be  with- 
out value  to  the  practical  engineer,  and  we  therefore  append 
them  here.  It  must  be  remembered  that  the  constructions 
hold  good  ONLY  for  end  spans  three-fourths  the  length  of  the 
others. 

I.  Beam  of  Uniform  Strength,  continuous  over  one 
Pier,  forming  two  equal  Spans,  subject  to  a  fixed  Load 
uniformly  distributed,  and  also  to  a  moving  'Load. 
Maximum  Moments.*  PI.  18,  Fig.  73. 

The  greatest  moment  at  the  pier  (positive)  will  be  when  both 
spans  are  fully  loaded. 

The  greatest  negative  moment  will  obtain  in  the  loaded  span 
when  the  other  span  bears  only  the  fixed  load. 

(A  moment  is  positive  when  the  upper  fibres  or  flanges  are 
extended,  negative  when  the  upper  flange  is  compressed.) 

Construction.  —  Let  A  B  C  be  the  beam.     On  A  B  draw  the 


parabola  whose  centre  ordinate   D  E  is   (p  +  m)  ^-,  and   on 

p  1? 
B  C  the  parabola  whose  centre  ordinate  G  F  is  "—-. 

i  •    * 

At  the  pier  B  erect  the  perpendicular  B  H  =  —  —  ^  —  —-., 

and  make  B  L  =          ^  0     -  —  .     Join  AH,  A  L,  and  L  C. 
\A 

*  "  Strains  in  Girders,  calculated  by  Formulas  and  Diagrams."    Humber. 
New  York  :  D.  Van  Nostrand,  publisher. 


164:  METHODS  OF  LOADING  [CHAP.  XI. 

Then  the  vertical  distances  between  the  parabolic  arc  A  E  B 
and  the  lines  A  H  and  A  L,  the  greatest  being  taken,  will  give 
the  maximum  moments,—  positive  in  the  first  case  and  negative 
in  the  last.  The  points  of  inflection  approach  as  near  the  pier 
as  K  and  recede  as  far  as  M. 

If  4r~  is  less  than  —  —  ,  the  beam  must  be  latched 

2i  \.2i 

down  at  the  abutments.  The  load  comes  on  from  the  left; 
p  and  m  are  the  loads  per  unit  of  length  of  the  permanent  or 
fixed  and  the  moving  or  live  loads. 

Shearing  Forces  (PI.  18,  Fig.  74).  —  The  maximum  shearing 
force  at  either  abutment  will  obtain  when  its  span  only  sus- 
tains the  moving  load.  The  maximum  shear  at  the  centre  pier 
will  obtain  when  both  spans  are  fully  loaded. 

Construction.  —  Lay  off  AC  =       (4:p  +  5m)  and  A  D  = 


o  (p  +  m).     At  B  lay  off  B  P  =  twice  A  D.     Take  a  point 

M  distant  -J  I  from  A,  and  join  D  and  F  to  M.  Draw  C  N 
parallel  to  D  M.  Sketch  in  a  curve  similar  to  that  dotted  in 
the  figure,  giving  an  additional  depth  to  the  ordinates  at  the 

point  of  minimum  shear  of  m  -.     Then  the  vertical  ordinates 

between  AB  and  COPF  may  be  considered  to  give  the 
maximum  shearing  force  for  either  span. 

II.  Beam  as  above  —  continuous  over  three  or  more 
Pier§.  li  =  end  spans.  I  =  the  other  spans. 

Moments. 

The  maximum  moment  (positive)  will  obtain  when  only  the 
two  adjacent  spans,  and  every  alternate  span  from  them,  are 
simultaneously  loaded  with  the  total  load  —  the  remaining  spans 
sustaining  only  the  fixed  load. 

The  maximum  moment  at  the  centre  of  any  span  will  obtain 
when  it  and  the  alternate  spans  from  it  are  fully  loaded  —  the 
remaining  spans  sustaining  only  the  fixed  load. 

Construction  (PL  18,  Fig.  75).  —  Let  A  B  C  be  part  of  the 
beam.  On  B  C  draw  the  parabola  whose  centre  ordinate  E  F 

=  —  ~  ,  and  on  A  B  the  parabola  whose  centre  ordi- 

nate C  D  =  (P  +  ™Wt     At  B  and  c  make  B  H  _  c  L  = 

8 


CHAP.  XI.]  CAUSING   MAXIMUM   STRAINS.  165 

Join  A  Hand  I,     Make  B  G  =  C  S  =  *l. 


o 

Join  A  G  and  S.  The  maximum  vertical  ordinates  between 
the  two  parabolas  and  the  lines  A  H,  A  G,  H  L,  and  G  S,  as 
shown  in  the  figure,  give  the  maximum  moments. 

The  points  I,  K  and  O,  P  or  M,  N  show  the  limits  of  devia- 
tion of  the  points  of  inflection. 

If  -^—  is  less  than  -  —  -  ,  the  beam  will  require  to  be 

2i  oA 

held  down  at  the  abutments. 

If  the  beam  be  continuous  for  three  spans  only,  I  in  the 

expression  for  B  H  =  —  (  —=-  +  —  )  must  have  a  value  given  to 


Shearing  Forces. 

The  maximum  shear  at  any  pier  (B  or  C)  will  obtain  simul- 
taneously with  the  maximum  moment  over  that  pier. 
Construction.—  PL  18,  Fig.  76. 
Let  A  B  C  be  part  of  the  beam.     First,  for  any  inner  span  as 

Z__.     At  B  and  C  erect  B  G  =  C  H  =  i&^J&A.     Make  B  D 


and  C  E  each  =  -  (p  +  m).     Join  D  and  E  to  midspan  F,  and 

2 

draw  G  K  and  H  K  parallel  to  D  F  and  F  E  respectively. 
Second,  for  either  end  span  as  \  —  .     At  B  erect  a  perpen- 

2  7  3 

dicular  =  —  -  (p+m),  which,  if  ^  =  -  I,  will  coincide  with  B  D. 

3  4: 

At  A  make  AL  =  -  B  D.  Join  D  and  L  to  M  distant  -  \  from  A. 
2  o 

Make  A  O  =|  (p  +  m)-  ^-2,  and  draw  ON  parallel  to  L  M. 

2i   '  32  1-±  • 

Sketch  in  curves  as  shown  by  the  dotted  curves  in  the  figure, 
giving  additional  depth  to  the  ordinates  there  of  —^  and  --ir- 

respectively. Then  the  vertical  distances  between  O  a  b  D  and 
A  B  give  the  maximum  shearing  forces  for  either  end  span,  and 
those  between  G  c  d  H  and  B  C,  the  shearing  forces  for  the  re- 
maining spans. 


166  METHODS  OF  LOADING  [CHAP.  XI. 

If  the  beam  be  continuous  for  three  spans  only,  B  G  and  C  H 


must  be  made  equal  to  (18  *  +  16  "*>*+"  (2-#  +  ?),  where 

oz  o  I  \  7       2i  ' 

L  =  —~.     Further,  the  value  given  to  B  D  for  the  inner  span 


3 
1O8.  JUethod  by  Resolution  of    Forces— Draw  Spans.— 

The  most  usual  cases  of  continuous  girders  which  occur  in  prac- 
tice are  draw  or  pivot  spans,  which  when  shut  must  be  consid- 
ered as  continuous  girders  of  two  spans.  The  graphical  method 
becomes  for  such  cases  short  and  easy  of  application.  In  the 
case  of  framed  structures  of  this  character,  it  may,  however,  be 
more  satisfactory  to  first  find  the  maximum  shearing  forces 
(Art.  104),  and  then  follow  the  reactions  thus  obtained  through 
the  structure  from  end  to  end  by  the  method  of  Arts.  8-13. 
As  a  check  upon  the  accuracy  of  the  work,  we  may  apply  the 
"  method  of  sections  "  referred  to  in  Art.  14.  In  either  case 
we  must,  of  course,  start  from  an  end  support  where  only  two 
pieces  intersect  and  the  moment  is  zero. 

Still  again,  we  may  find  the  reactions  by  calculation,  and 
then  apply  the  method  of  Arts.  8-13.  In  the  case  of  two  spans 
only,  the  formulae  for  the  reactions  are  sufficiently  simple,  and 
the  ready  and  accurate  determination  of  the  strains  offers,  there- 
fore, no  difficulty. 

We  shall  give  here,  therefore,  the  analytical  formulae  requi- 
site for  our  purpose,  referring  the  reader  to  treatises  upon  the 
subject  for  their  demonstration.* 

Formulae  for  Reactions. —  Continuous  Girder  of  two  un- 
equal /Spans,  I  and  n  I.  1st.  Concentrated  weight  P,  in  first 
span  Z,  distant  /3  from  left  end  support.  Reaction  at  left  end 
support : 

P 


Reaction  at  middle  support : 


*  Bresse — La  Flexion  et  la  Resistance,  and  Cours  de  Mecanique  Appliquee. 
Weyrauch — Theorie  der  Trdger.  CoUignon — Theorie  Elementaire  des  Poutres 
Droites,  etc.  Also  Supplement  to  Chap.  XIII. 


CHAP.  XI.]  CAUSING  MAXIMUM   STRAINS.  167 

Reaction  at  right  support : 

0=         _P f-^4-^1 

2tt(l  +  7l)    L        J  "*>_!• 

2d.  Uniformly  distributed  load  extending  to  a  distance  /3 
from  left  support.     Load  per  unit  of  length  =p. 


o= 


For  two  equal  spans  we  have  only  to  make  n  =  1  in  the 
above  equations.  For  a  uniform  load  over  whole  span  /3  =  L 

From  the  above  formulae  we  can  find  the  reactions  for  any 
case,  and  then  proceed  as  indicated  above. 

109.  By  means  of  the  graphical  method,  as  we  have  now 
seen,  we  are  enabled  to  solve  completely  the  problem  of  the 
continuous  girder,  and  that  too  without  the  aid  of  analytical 
formulae,  tables,  or  tedious  computation.  The  method  can  also 
be  applied  to  continuous  girders  of  variable  cross-section,  or  of 
uniform  strength.  We  shall  not,  however,  proceed  further 
with  the  development  of  the  method  in  this  direction.  The 
preceding  will,  we  think,  be  found  to  contain  all  that  is  practi- 
cally serviceable.  For  the  application  of  the  method  to  girders 
of  variable  cross-section,  we  refer  the  reader  to  Winlder  —  "  Der 
Bruckenbau"  Wien,  1873  —  where  will  be  found  a  thorough 
presentation  of  the  subject,  both  analytically  and  graphically, 
to  which  we  are  greatly  indebted  in  the  preparation  of  the 
preceding  pages.  Plates  17,  19  and  20,  are,  with  but  few  alter- 
ations, reproduced  from  that  work. 

*  These  formal  EG  are  demonstrated  in  Van  Nostranffs  Eng.  Mag.,  July, 
1875. 


168  GRAPHIC    AND    ANALYTIC  [CHAP.  XH. 


CHAPTEE   XII. 

CONTINUOUS     GIRDER     (CONTINUED) COMBINATION      OF      GRAPHICAL 

AND    ANALYTICAL   METHODS. 

1 1O.  In  the  present  chapter  we  shall  develop  a  method  for 
the  solution  of  continuous  girders  not  purely  graphical,  but 
based  upon  the  method  of  resolution  of  forces  illustrated  in 
Arts.  8-13,  together  with  well-known  analytical  results,  which 
method  for  accuracy,  simplicity,  and  ease  of  application  will, 
we  think,  be  found  superior  to  any  hitherto  proposed.      The 
method  is,  of  course,  applicable  only  to  framed  structures,  but 
for  such  cases  is  the  most  satisfactory  of  any  with  which  we  are 
acquainted. 

111.  The  Inflection  Points  being  known,  the  Shearing 
Forces  and  Moments  at  the  Support*    can,  by  a  simple 
construction,  be  easily  determined. — ~Lst.    Loaded  Span — 
Fig.  77,  PL  21.— Thus  in  the  span  B  C  =  I,  let  the  distance  of 
the  weight  P  from  the  left  support  be  a,  and  let  i  and  i'  be  the 
distances  of  the  inflection  points  from  B  and  C  respectively. 
Then  if  through  any  point  P  of  the  weight  we  draw  lines,  as 
P  D,  P  E,  through  i  and  i',  intersecting  the  verticals  at  B  and  C 
in  the  points  D  and  E,  the  vertical  ordinates  between  these  lines 
and  B  C  will  be  proportional  to  the  moments.     For,  as  we  see 
from  the  force  polygon,  the  equilibrium  polygon  must  consist 
of  two  lines  as  D  P,  P  E,  parallel  to  O  0  and  O  1,  and  because 
of  the  moments  at  the  ends,  the  closing  line  D  E  is  shifted  to 
B  C  (Art.  23).     Since  the  moments  at  the  points  of  inflection 
are  zero^  the  ordinates  to  P  D  and  P  E  to  pole  distance  H  will 
give  the  moments.     Now  the  points  of  inflection  being  known, 
and  P  J3  and  P  E  drawn,  we  can  easily  find  the  pole  distance 
H  and  the  shearing  forces  L  0  and  1  L  by  laying  off  P  verti- 
cally, and  drawing  from  its  extremities  lines  parallel  to  P  D 
and  P  E  intersecting  in  O.     A  perpendicular  through  O  upon 
0  1  gives  H  and  the  reactions  L  0  and  1  L.     In  other  words, 
we  have  simply  to  decompose  P  along  P  D  and  P  E. 


CHAP.  XII.]  METHODS    COMBINED.  169 

The  construction,  then,  is  simply  as  follows :  Take  any  point 
on  the  direction  of  P,  and  draw  P  D,  P  E  through  the  points 
of  inflection.  Lay  oft  P  to  the  scale  of  force  as  A  P,  and  draw 
A  O  parallel  to  P  E  or  P  D.  We  have  thus  the  pole  distance 
H,  and  the  shearing  forces  P  H  and  H  A  at  B  and  C. 

B  D  or  C  E  to  the  scale  of  distance,  multiplied  by  H  to  the 
scale  of  force,  give  the  moments  at  B  and  C.  That  is,  D  and 
E  may  be  regarded  as  the  points  of  application  for  H.  The 
forces  along  P  D  and  P  E  considered  as  acting  at  these  points 
are  held  in  equilibrium  by  the  reactions  P  H  and  H  A  =  L  0 
and  1  L  and  H.  Since  H,  acting  as  indicated  in  the  figure 
with  the  lever  arm  B  D  or  C  E,  causes  tension  in  the  upper 
fibres,  the  moments  at  B  and  C  are  positive. 

2d.  Unloaded  Span—Fig.  78,  PL  21. 

As  we  have  already  seen  in  Art.  93,  the  inflection  points  in 
the  unloaded  spans  are  independent  of  the  load,  and  are  found 
by  the  simple  construction  there  given  for  the  "fixed points" 
Since,  each  fixed  point  lies  within  the  outer  third  of  the  span, 
we  have  in  Fig.  78  the  broken  line  a  b  c,  referred  to  in  Art.  94, 
where  the  moments  are  alternately  positive  and  negative,  and 
increase  from  the  end,  so  that  any  one  is  more  than  twice  the 
preceding.  Lines  drawn  parallel  to  these  lines  in  the  force 
polygon,  cut  off  from  the  force  line  the  reactions  at  the  sup- 
ports. Thus,  in  •  Fig.  78,  c  b  in  the  force  polygon  gives  the 
reaction  at  D,  a  b  the  reaction  at  C,  and  if  B  were  an  end  sup- 
port— that  is,  if  b  a  went  through  B — a  H  would  be  the  re 
action  at  B.  For  the  resultant  shear  at  D,  we  should  then  have 
aH  —  ab+-cb  =  Hc.  So  for  any  number  of  spans  ;  the  in- 
flection points  in  the  loaded  span  being  known,  we  can  easily 
find  the  fixed  or  inflection  points  in  the  other  spans,  which  are 
independent  of  the  load,  and  depend  only  on  the  length  of 
these  spans.  Then  draw  the  broken  line  a  b  c  P  d.  Then  find 
the  pole  distance  H  by  laying  off  c  P  =  P  to  scale,  and  draw- 
ing c  O  parallel  to  c  P,  and  through  the  point  O  thus  deter- 
mined drawing  H  O.  Then  find  the  reactions  at  the  other 
supports,  or  the  shear  at  any  support,  by  lines  in  the  force 
polygon  parallel  to  a  b,  b  c,  etc.  Thus  the  shear  at  B  is  the 
distance  H  a  cut  off  by  H  and  O  a  parallel  to  a  b.  Since  the 
shear  at  D  is  plus  and  alternates  from  D,  we  have  at  B  the 
shear  +  H  a.  The  shear  at  C  is  —  H  b ;  at  D,  H-  H  c,  etc.  O  c 
being  parallel  to  PC;  O  a,  to  b  a ;  O  b,  to  c  b,  etc. 


170  GRAPHIC   AND   ANALYTIC  [CHAP.  XII.  * 

112.  Inflection  Verticals,  —  Draw  a  line  from  P  through 
the  support  D,  and  through  its  intersection  with  cb  draw  a 
vertical  3  (Fig.  78).  This  vertical  we  call  the  inflection  ver- 
tical. 

The  equation  of  the  line  c  ft  is 


where  ml  =  D  C,  n  I  =  C  D,  i  =  Ci'.     The  origin  being  at  D. 
For  the  line      C, 


where  ^  =  D  i±. 

If  in  this  last  equation  we  make  x  =  a,  we  have  for  the  or- 
dinate  at  P, 

1ft!  fa  —  a) 

h 
and  hence  for  the  line  P  D, 

mt  (*,  —  a) 
y  =  —  *;  -  -x. 


For  the  intersection  of  P  D  with  b  c  then 

m±  m±  (^  —  a) 

—  ^  -  .  x  +  m±  =  —  ^  -  -  x. 
nl  —  v  i^a 

Hence 


-    /.  \     /        7  -\  •  «... 

(^  —  a)  (n  I  —  i)  —  \  a 

We  see  at  once  that  the  value  of  x  is  independent  of  ml  or 
B  C,  hence  the  intersection  of  P  D  and  c  b  lies  always  in  the 
same  vertical,  whatever  be  the  position  ofPC.  In  other  words, 
if  the  three  sides  of  a  triangle  pass  always  through  three  fixed 
points  (a1,  D,  ^),  and  two  of  the  angles  (P  and  c)  be  always  in 
the  same  verticals,  the  third  angle  must  also  always  lie  in  the 
same  vertical. 

For  the  distance  of  the  inflection  vertical  on  the  other  side 
of  the  loaded  span  (beyond  E),  we  have  similarly 


(I  —  a)  («B  +  4) 

where  I  is  the  loaded  span  and  iB  the  distance  of  the  inflection 
point  to  the  right  of  E. 

Equations  (1)  and  (2)  give  the  distances  of  the  inflection  ver- 
ticals from  the  supports  D  and  E. 


CHAP.  XII.]  METHODS    COMBINED.  171 

113.  Beam  fixed  horizontally  at  both  ends—  Supports 
011  level.  —  Consider  the  span  D  E  (Fig.  78)  as  fixed  at  the  sup- 
ports so  that  the  tangent  to  the  deflection  curves  at  D  and  E  is 
always  horizontal.  Conceive  the  span  prolonged  right  and  left 
beyond  the  supports  a  distance  equal  to  the  span  I.  It  is  re- 
quired to  find  the  position  of  the  inflection  verticals. 

From  equation  (1)  of  the  preceding  Art.  we  have,  since 
n  =  1,  i  =  0, 

ii  a  \ 

t/U  • 


""•  7       -w  •  • 

tj  —  a)  l  —  ^la/ 
and  from  equation  (2),  since  is  —  I, 

i%  I  (I  —  a) 


x  = 


Now  for  a  beam  fixed  at  the  ends  the  distances  of  the  points 
of  inflection  are 

a  I  I  (I  —  a) 


Substituting  these   values   in   the  equations   above,  we   have 

x  =  —  -  and  x  =  4-  o-     That  is,  the  position  of  the  inflection 
6  o 

verticals  is  in  this  case  independent  of  the  load,  and  always 

equal  to  ~  from  the  supports* 
o 

This  remarkable  property  of  the  beam  fixed  at  both  ends 
enables  us  to  find  the  inflection  points  by  a  construction  similar 
to  that  for  the  fixed  points  in  the  unloaded  spans,  as  given  in 
Art.  86. 

Thus  we  have  simply  to  draw  from  C  distant  I  from  A  (PI. 
21,  Fig.  79)  a  line  in  any  convenient  direction,  as  C  ~b  intersect- 

ing the  inflection  vertical  I,  which  is  distant  from  A,  -  Z,  at  a. 

Through  a  and  the  fixed  end  support  A  draw  a  line  to  inter- 
section with  the  weight  P.  Then  draw  P  b.  The  intersection 
it  of  this  last  line  with  A  B  is  the  inflection  point.  A  similar 
construction  gives  i%. 

We  can  now  find  the  reactions  and  moments.     Thus  H  O  to 

*  This  important  result,  which  renders  possible  a  complete  graphical  solution 
of  this  case,  has,  so  far  as  we  are  aware,  never  before  been  published. 


172  GRAPHIC   AND   ANALYTIC  [CHAP.  XII. 

the  scale  of  force,  multiplied  by  A  b  to  the  scale  of  distance, 
gives  the  moment  at  A,  while  H  G  is  the  reaction  at  A  (Fig.  79). 
114.  Beam  fixed  at  both  ends— Example. — Since  when 
the  points  of  inflection  are  once  determined,  we  may  draw  P  b 
or  P  c  at  any  inclination  (Fig.  79),  provided  we  afterwards  find 
the  corresponding  pole  distance  HO;  if  A  b  or  B  c  be  made 
equal  to  the  height  of  the  truss,  H  O  will  ~be  the  strain  in  the 
upper  or  lower  flange  at  the  wall  (the  flange  in  question  being 
always  that  for  which  there  is  no  diagonal  at  its  union  with  the 
wall).  Thus  in  PL  21,  Fig.  80,  we  lay  off  D  E  =  I,  draw  the 

vertical  I  at  ~  I  from  D,  and  for  the  given  position  of  the  load 

o 

P  find  the  inflection  point  4  by  the  preceding  Art.  A  similar 
construction  on  the  other  side  gives  %.  Now  laying  off  P  M 
equal  by  scale  to  the  weight  P,  and  decomposing  it  along  P  D 
and  P  C,  we  find  O  H  the  pole  distance  which  to  the  scale  of 
force  will  give  directly  the  strain  in  the  lower  flange  B  m  at 
the  wall,  provided  P  D  is  made  to  pass  through  the  intersection 
of  the  upper  flange  with  the  wall.  If  the  triangulation  were 
reversed,  O  H  would  be  the  strain  in  the  upper  flange  at  the 
wall.  In  any  case  it  is  the  strain  in  that  flange  at  whose  junc- 
tion with  the  wall  there  is  no  diagonal. 

The  reaction  at  D  is  also  H  M,  at  C  it  is  P  H.  Lay  off  then 
B  B'  in  Fig.  80  (a)  equal  to  P  M,  and  make  B  A  =  H  M  and 
A  B'  =  P  H.  Now  draw  m  B  parallel  to  O  H  and  m  A  paral- 
lel to  O  M,  and  produce  both  lines  to  intersection  at  m.  Then 
m  B  to  scale  of  force  is  evidently  the  strain  in  the  lower  end 
flange  at  the  wall.  We  assume  the  following  notation.* 

Let  A  represent  all  the  space  above  the  girder,  B  all  the  space 
below,  and  a  h  c  d7  etc.,  the  spaces  within  the  girder  included 
by  the  flanges  and  diagonals.  Then,  for  instance,  A  b  is  the  first 
upper  flange  at  the  left,  B  a  the  first  below ;  a  b  the  first  diag- 
onal at  the  left,  and  so  on. 

Now  draw  in  Fig.  80  (#),  m  I  and  A  I  parallel  to  the  corre- 
sponding lines  in  the  frame,  and  we  have  at  once  the  strains  in 
these  pieces  to  scale.  Following  round  the  triangle  according 
to  our  rule  (Arts.  8-13)  from  m  to  A,  A  to  I  and  I  «to  m,  we 

*  See  an  excellent  little  treatise  on  "  Economics  of  Construction  in  Relation 
to  Framed  Structures,"  by  R.  H.  Bow,  to  whom  this  method  of  notation  is 
due. 


CHAP.  XII.]  METHODS    COMBINED.  173 

find  A  I  tension  and  I  m  tension.  [The  strains  in  the  upper 
flanges  must  always  be  tension,  since  the  moments  at  the  sup- 
ports for  loaded  span  are  always  positive.]  Moreover,  the  Fig. 
thus  far  shows  that  A  I,  I  m  and  m  B  are  in  equilibrium  with 
the  shear  B  A  =  H  M,  as  evidently  should  be  the  case ;  hence 
the  strain  in  B  m  is  compressive. 

We  have  thus  the  strains  in  the  three  pieces  at  the  right,  and 
can  proceed  from  these  to  find  all  the  others.  Thus  the  strains 
in  B  k  and  k  I  are  in  equilibrium  with  m  I  and  B  m.  Lines 
parallel  to  B  It.  and  k  I,  therefore,  which  close  the  polygon  com- 
menced by  B  m  and  m  I,  give  us  the  strains  in  B  k  and  k  I. 
Observe  that  the  line  I  k  crosses  A  B,  thereby  making  B  k  op- 
posite in  direction,  consequently  in  strain  from  B  m.  This  may 
also  be  seen  by  following  round  the  triangle  m  1 k  B,  remem- 
bering that,  as  m  I  is  always  found  to  be  in  tension,  it  must  act 
away  from  the  new  apex,  that  is,  from  m  to  I.  We  thus  find 
k  I  in  compression,  and  k  B  acting  away  from  this  apex,  or  in 
tension,  therefore  of  opposite  strain  from  the  preceding  flange 
B  m,  which,  as  we  have  seen,  is  in  compression.  The  reason  is 
obvious.  The  inflection  point  i%  falls  in  the  flange  B  k.  If  the 
beam  wrere  solid,  the  strain  at  i2  would  be  zero ;  to  the  right  of 
4  we  should  have  compression,  to  the  left,  tension.  In  the 
framed  structure  the  strains  can  only  change  at  the  vertices. 
The  crossing  of  A  B  by  I  k  indicates  such  change,  and  B  k  gives 
its  amount  by  scale. 

Now  taking  the  upper  apex,  we  have  here  A  I  and  I  k  in 
equilibrium  with  k  h  and  A  h.  As  we  already  know,  k  I  is  in 
compression.  We  must,  therefore,  now  take  it  acting  from  k  to 
I,  and  following  round  the  triangle  we  find  A  h  compression, 
and  h  k  tension.  From  h  on,  the  traverses  between  A  h  and 
B  k  produced  towards  the  right  [Fig.  80  (&)]  will  give  the  di- 
agonals, while  the  upper  and  lower  flanges  will  be  given  by  the 
distances  to  them  from  A  and  B  respectively,  until  we  arrive 
at  the  weight  P.  Observe  the  influence  of  the  weight.  We 
have  k  h  and  B  k  in  equilibrium  with  h  g  and  g  B,  and  also  the 
weight  P  =  B'  B.  We  must  take,  therefore,  A  B'  =  P  H,  and 
then  draw  h  g  and  B'  g.  Distances  to  the  right  of  B'  along 
B'<7  are  compressive  lower  flanges,  to  the  left,  tensile;  while  to 
the  right  of  A  we  have  compressive  upper,  and  to  the  left  of  A 
tensile  upper  flanges.  The  two  diagonals  at  the  weight  k  h  and 
h  g  are  in  tension.  From  h  g  on,  the  diagonals  are  alternately 


174  GRAPHIC  AND  ANALYTIC  [CHAP  XII. 

tension  and  compression.  Moreover,  the  diagonal  e  d  passes 
through  A,  that  is  A  d  is  zero.  The  weight  P  causes  no  strain 
in  A  d,  and  for  this  one  position  of  P,  A  d  might  be  omitted 
from  the  structure.  The  reason  is  again  obvious.  The  point 
of  inflection  i±  coincides  with  the  apex  under  A  d.  Since  at 
^  the  moment  of  rupture  is  zero,  if  the  flange  A.~d  were  cut 
there  would  be  no  tendency  to  motion.  We  have  at  ?'t  the 
shearing  force  only,  A  B'  giving  the  strains  in  the  diagonal 
e  d  and  d  c.  The  upper  flange  A  £',  we  see  again,  is  in  tension, 
which  is  also  shown  by  its  lying  to  the  left  of  A. 

Thus  we  have  the  strains  in  every  piece  by  a  very  simple 
construction  for  any  position  of  P,  without  any  calculation  what- 
ever. The  method  in  this  case  is  purely  graphical.  We  have 
only  to  find  the  points  of  inflection  and  then  proceed  as  above. 

Fig.  80  (b)  gives  the  strains  for  the  same  girder  and  position 
of  weight  P,  merely  supported  at  the  ends.  For  this  case  P  D 
in  Fig.  80  not  only  passes  through  D,  but  P  O  also  passes  through 
the  upper  left-hand  corner  at  C.  Hence  A  B  will  be  less  than 
H  M,  and  A  B'  greater  than  P  H.  Moreover,  the  end  lower 
flanges  B  a  and  B  m  no  longer  act,  and  must  be  removed. 
Starting  now  with  the  reaction  B'  A  [Fig.  80  (£)],  we  go  along 
to  the  weight,  from  which  point  at  h  k  we  go  ~back  towards  the 
force  line,  and  the  reactions  are  such  that  the  last  diagonal 
must  pass  exactly  through  B,  just  as  in  Fig.  (a)  e  d  passed 
through  A,  because  the  points  of  inflection  or  zero  moments  are 
now  at  the  ends  C  and  D.  A  careful  comparison  and  study  of 
the  two  cases  and  their  points  of  difference  will  be  advanta- 
geous to  the  reader. 

115.  Counterbracing. — The  objection  may  arise  that  the 
above  method  applies  only  to  a  system  of  bracing  such  as  rep- 
resented in  the  Fig.,  where  the  diagonals  take  both  compres- 
si ve  and  tensile  strains.  In  case,  as  in  the  Howe  or  Pratt  Truss, 
for  instance,  we  had  vertical  pieces  as  also  two  diagonals  in 
each  panel,  then  the  strain  in  any  diagonal  as  m  I  and  flange  as 
B  m,  even  if  found,  are  apparently  in  equilibrium  with  three 
pieces,  viz.,  Jcl,~Bk  and  a  vertical  strut  or  tie  at  the  intersec- 
tion of  these  pieces.  Hence,  having  only  two  known  strains 
and  three  to  be  determined,  the  method  would  seem  to  fail,  as 
any  number  of  polygons  may  be  constructed  with  sides  par- 
allel to  the  forces,  and  hence  the  problem  is  indeterminate 
(Art.  9). 


CHAP.  XII.]  METHODS    COMBINED.  175 

jSTow  in  any  framed  structure  of  the  above  kind,  the  counter 
ties  are  inserted  to  prevent  the  deforming  action  of  the  Tolling 
load  only.  For  the  dead  load  but  one  system  of  triangulation 
is  required,  and  the  strains  in  every  piece  due  to  this  dead  load 
can  therefore  easily  be  determined. 

We  have  then  only  to  determine  the  strains  in  the  same  pieces 
due  to  the  rolling  load  also.  If  now  in  any  diagonal  the  strain 
due  to  this  rolling  load  exceeds  the  constant  strain  due  to  the 

O 

dead  load,  and  is  of  opposite  character,  and  if  the  diagonal  is 
to  be  so  constructed  as  to  take  biit  one  kind  of  strain,  then  a 
counter  diagonal  must  be  inserted  in  that  panel,  and  propor- 
tioned to  this  excess  of  strain  only.  For  instance,  if  a  diagonal 
takes  only  the  compressive  strain  (a  condition  which  is  easily 
secured  in  practice)  due  to  the  dead  load,  and  the  live  load 
would  cause  in  that  diagonal  a  tensile  strain,  then  the  excess  of 
this  tensile  strain  over  the  constant  compressive  strain  due  to 
the  dead  load  must  be  resisted  by  a  counter  diagonal,  which 
also  takes  compressive  strain  only.  The  method  is  precisely 
the  same  as  by  calculation  (see  Stoney  and  other  authors  011 
the  subject),  and  we  only  notice  the  point  here,  as  in  all  our 
examples  we  have  taken  a  single  system  of  triangulation  only — 
a  system  which,  we  may  here  remark  in  passing,  has  many  ad- 
vantages, and  is  worthy  of  more  general  attention  *  than  it  has 
hitherto  obtained. 

[See  also  on  this  point  Art.  10  of  Appendix.] 
116.  Beam  fixed  horizontally  at  one  end,  supported  at 
the  other— Supports  on  Level. — In   this  case,  equation  (1), 
Art.  112,  becomes  for  left  end  fixed,  since  n  =  1,  i%  =  0,  i  —  0, 

i-i  a  I 


(^ — a)  l — ^\  a 

But  for  this  case  the  distance  of  the  point  of  inflection  from 
the  fixed  end  is 

.  _        (2  I— a)  la     , 


*  See  "  A  Treatise  on  Bracing."  By  R.  H.  Bow.  D.  Van  Nostrand,  pub- 
lisher. 

f  The  values  of  the  distance  of  the  inflection  points  which  we  assume  above 
as  known,  may  easily  be  deduced  by  the  theory  of  elasticity.  See  Supplement 
to  Chap.  VTL,  Arts.  16  and  19.  See  Wood,  Strength  of  Materials ;  Bresse, 
Mecanique  Appliquce  ;  or  other  treatises  on  the  subject. 


176  GRAPHIC  AND  ANALYTIC  [CHAP.  XII. 

Inserting  this  value  of  ^  in  the  value  for  x  above,  we  have 

x  -  _  l  (2  t-a) 
5  I— a 

for  the  distance  of  the  inflection  vertical  to  the  left  of  the  left- 
hand  support,  which  is  supposed  fixed. 

Now  this  is  the  equation  of  an  hyperbola,  as  shown  in  PL  21, 
Fig.  81,  whose  vertex  is  at/J  the  distance  Ay  being  2  I,  whose 
assymptotes  are  respectively  parallel  and  perpendicular  to  the 
span,  the  perpendicular  distance  of  E  above  the  span  A  B  be- 

5  9 

ing  1  I,  and  which  intersects  A  B  at  ^  I  from  A.     The  ordinate 
2i  5 

d  e,  A  d  being  equal  to  Z,  is  -  1.     The  diameter  passes  through 

3 

E  and  f,  and  E  f  is,  therefore,  the  semi-transverse  axis.  The 
hyperbola  can,  therefore,  be  easily  constructed.  We  need  only 
to  construct  that  portion  between  A  B  and  the  point  e. 

The  construction  for  the  point  of  inflection  i±  is,  therefore, 
simply  as  follows : 

Lay  off  A  k  vertically  upwards  and  equal  to  the  distance  of 
the  weight  P  from  A,  and  draw  the  horizontal  h  b  to  intersection 
b  with  the  curve.  Now  make  A  a  =  I  and  draw  a  b  to  inter- 
section c.  Draw  b  A  to  intersection  P  with  weight,  and  then 
P  c  intersects  A  B  at  the  point  of  inflection  ilm  Decomposing 
P  along  P  B  and  P  c,  as  in  Art.  114,  we  have  at  once  the  re- 
actions at  A  and  B.  Here  also  we  see  that,  by  a  construction 
purely  graphic  and  abundantly  exact,  we  can  find  the  inflection 
point  and  the  reactions. 

The  method  detailed  in  Art.  114  can  then  be  applied  to  de- 
termine the  various  strains  in  the  different  pieces.  It  is  un- 
necessary to  give  an  example,  as  the  process  is  precisely  similar. 
We  have  simply  in  this  case  to  start  with  the  reaction  at  the 
free  end  B  and  follow  it  through.  Observe  only  that,  as  this 
reaction  must  be  less  than  for  a  girder  with  free  ends  for  the 
same  position  of  P,  the  point  h  will  lie  nearer  the  force  line 
B' AB  (Fig.  SO,  &),  hence  Im  will  not  pass  exactly  through  B, 
but  will  lie  to  the  right  of  it,  giving  thus  a  reversal  of  strain  in 
the  flanges,  as  by  reason  of  the  inflection  point  should  be  the 
case. 

Instead  of  constructing  tho  hyperbola,  we  may  calculate  its 
ordinates  from  the  equation  for  x  above,  for  different  values 
of  a. 


CHAP.  XII.]  METHODS    COMBINED.  177 

Thus,  for 

x=—OAl  %=-0.3SSl  x=-0.375l  o?=— 0.35YZ  aj=— 0.333  J 

This  will  be  sufficient  to  construct  the  curve  in  any  given 
case.  The  inflection  vertical  moves,  therefore,  between  the 
narrow  limits  of  x  =  f  I  and  %  =  %  I,  or  within  J^th  of  the  span, 
as  the  load  passes  from  A  to  B. 

Inasmuch  as  all  that  is  needed  for  the  determination  of  the 
strains  in  the  various  pieces  are  the  reactions  at  the  supports, 
and  (for  girder  fixed  at  both  ends)  the  moments  at  the  supports 
also,  and  as  the  formulae  for  the  two  cases  above  are  very  sim- 
ple, we  may  determine  these  quantities  at  once  by  interpolation 
of  the  given  distance  of  the  weight  P  in  the  formulae,  and  then 
apply  the  graphical  method  for  the  strains,  as  illustrated  in 
Art.  114. 

Thus,  for  a  horizontal  beam  fixed  at  both  ends,  we  have  for 
the  moment  at  the  left  support  A, 

-  —  V 

At  the  right  support  B,  .. 

MB  =  ^  (I  -  a). 
For  the  reaction  at  the  left, 


For  the  reaction  at  the  right, 

RB  =  ?  (3  a2l  -  2  a3). 

In  the  case  of  a  horizontal  beam  fixed  at  left  end  and  merely 
Besting  upon  the  right  support,  we  have 

MA  =        (3  a^l  -  2  a  1?  -  a*\  MB  =  0, 


a  being  always  the  distance  of  the  weight  P  from  the  left. 
These  formulae  are  simple,  and  easily  applied  to  any  case. 

We  may  also  observe  that  in  Figs.  79  and  81  the  ordinates  to 
the  lines  P  5,  P  c,  and  P  c,  P  B,  from  A  B,  are  proportional  to 
the  moments  (Art.  110).  These  ordinates  to  the  scale  of  dis- 


178  GRAPHIC    AND  ANALYTIC  [cilAP.  XTT. 

tance,  multiplied  by  the  pole  distance  to  scale  of  force,  give  the 
moments  at  any  point.  '  Onr  construction,  therefore,  gives  the 
moments  also  at  every  point,  and  we  may  thus  check  the  re- 
sults obtained  by  Art.  114  by  the  results  obtained  by  the 
method  of  moments. 

117.  Approximate  Construction.  —  It  will  be  readily  seen 
that  the  portion  of  the  hyperbola  in  Fig.  81,  PI.  21,  needed  for 
our  construction,  is  nearly  straight.     In  most  cases  it  will  be 
practically  exact  enough  to  lay  off  f  I  to  the  left  of  A,  and  -J  I 
also  to  the  left  of  A  at  a  vertical  distance  equal  to  Z,  and  join 
the  two  points  thus  obtained  by  a  straight  line.     This  line  can 
be  taken  instead  of  the  curve,  and  the  construction  is  then  the 
same  as  above.     The  error  due  to  thus  considering  the  curve 
as  a  straight  line  is  greatest  for  a  weight  in  the  middle  of  the 
span,  where  it  does  not  exceed  To^th  of  the  span  for  the  posi- 
tion of  the  inflection  vertical,  and  diminishes  from  the  centre 
both  ways. 

118.  Girder  continuous   over  three  Level    Supports- 
Draw  Spans.  —  This   case   is   perhaps   of  the  most  frequent 
practical  occurrence,  and  an  accurate  and  simple  method  of 
solution  is  therefore  very  desirable. 

In  the  first  place,  the  formulae  for  the  reactions  are  very 
simple  and  easy  of  application.  Thus,  for  left  end  support  A, 
the  load  being  in  the  second  span,  or  to  the  right  of  the  middle 
support  B, 

RA  =  j?(3«^-2«?-«3); 

•*•  ^ 

for  the  reaction  at  middle  support, 


for  reaction  at  right  end  C, 


where  a  is  always  the  distance  of  the  weight  P  from  the  mid- 
dle support.*  We  are  therefore  already  in  a  position  to  solve 
completely  the  case  under  consideration.  We  have  only  to 

*  As  already  remarked,  the  development  of  the  formulas  assumed  in  this 
chapter  must  be  sought  for  in  special  treatises  on  the  subject.  We  assume 
them  as  known,  and  then  apply  them  graphically  as  above. 

See  also  Supplement  to  Chap.  XIII. 


CHAP.  Xn.]  METHODS    COMBINED.  179 

find  the  reactions  and  follow  them  through  by  the  method  of 
Art.  114. 

From  the  above  reactions  we  can,  however,  easily  determine 
the  distance  of  the  inflection  point.  This  will,  of  course,  be 
found  only  in  the  loaded  span,  at  a  distance  from  the  middle 
support. 


VU     -     ~^      TQ        \         ^T  7  O      '  -"      v« 

4&&+  2  al—  a2 

We  can  find  the  values  of  x  corresponding  to  different  values 
of  a,  and  thus  plot  the  curve  for  the  inflection  points.  Thus, 
for 

a  —  :0        a  =  ^l          a  =  $l          a  =  %l          a  =  I 
x  =  0        cc  —  -fa  I        x  =  -f^l        x  =  ^l        x  =  \l. 

This  curve  being  drawn  for  any  particular  case,  we  can 
easily  find  the  position  of  the  inflection  point  for  any  given 
value  of  #,  and  hence  the  reactions,  and  then  find  the  strains  in 
the  various  pieces. 

Thus,  in  PL  21,  Fig.  82,  the  curve  B  e  d  being  drawn,  we 
can  at  once  find  the  inflection  point  i  for  any  position  a  of 
the  weight  P.  We  have  simply  to  make  B  J  —  a  and  draw  5  e. 
b  e  is  the  distance  of  the  point  of  inflection  from  B.  We  can 
now,  as  explained  above,  draw  any  line  as  P  i,  and  then  P  C 
and  A  h.  The  ordinates  to  the  broken  line  A  h  P  C  from  A  C, 
to  the  scale  of  distance,  multiplied  by  the  pole  distance  H  to 
scale  of  force,  will  give  the  moments  at  any  point.  Moreover, 
H  E  is  the  shear  at  B.  E  a  is  the  reaction  at  B,  H  a  the  re- 
action at  A,  and  H  P  the  reaction  at  C.  The  reactions  at  B 
and  C  are,  of  course,  positive  or  upwards,  that  at  A  negative 
or  downwards.  Hence  E  «  —  H  #  -f-  H  P  =  P,  as  should  be. 

The  value  of  x  for  the  inflection  vertical  is  by  Art.  112 

i  a  I 

X  =  -T-.  -  r-j  -  ;  —  j 

(i  —  a)  I  —  ^  a 
or,  substituting  the  value  of  i  above, 

a  I  (2  I  -  a) 


Since,  therefore,  in  this  case  the  value  of  x  is  no  simpler  than 
that  for  i  given  above,  it  will  be  preferable  to  plot  the  first 
curve  directly  as  represented  in  Fig.  82. 

119.  Approximate  Construction.— In  practice  it  will  be 


180  GRAPHIC    AND   ANALYTIC  [CHAP.  XII. 

found  abundantly  accurate  to  assume  the  curve  for  i  between 
the  required  limits,  as  a  parabola  whose  equation  is  i  =  x  = 

a  ^     — -.     The  greatest  error  for  a  =  -  will   then  be  about 

5   t  4: 

—  of  the  span,  and  decreases  both  ways  to  a  =  o  and  a  =  -. 
100  *  2 

From  a  =  —  to  a  =  I  the  parabola  coincides  closely  with  the 

2 

q  -i 

true  curve.     The  difference  for  a  =  —  I  is  only  -^^-l,  and  we 

4  500 

have,  therefore,  a  very  simple  practical  construction  for  both 
reactions  and  moments.  We  have  only  (Fig.  82)  to  erect  a 
vertical  at  the  centre  support  B  and  make  it  equal  to  I,  and 
then  construct  a  parabola  passing  through  B  whose  ordinate 

c  d  =  -  I.     The  horizontal  ordinates  to  this  parabola  for  any 
5 

vertical  value  of  &,  give  the  distance  out  from  B  of  the  inflection 
points.  For  the  load  in  first  span  A  B,  of  course  this  parabola 
lies  on  the  other  side  of  B  c,  and  b  e  is  laid  off  to  the  left.  The 
remainder  of  the  construction  is  as  in  Art.  118  for  the  reactions 
and  moments.  When  great  accuracy  is  required,  we  can  find 
the  reactions  from  the  equations  of  Art.  118.  In  any  case,  the 
reactions  being  given,  we  can  follow  them  through  the  structure 
by  the  method  of  Art.  114,  and  thus  determine  the  strains  in 
every  piece  due  to  every  position  of  each  apex  load.  A  tabu- 
lation of  these  strains  will  then  give  by  inspection  the  maximum 
strain  in  any  piece  due  to  the  live  load.  All  the  weights  taken 
as  acting  simultaneously  will  then  give  the  strains  due  to  uni- 
form total  live  load.  The  strains  due  to  dead  load  will  be 
multiples  or  sub-multiples  of  these.  Thus  if  total  live  load 
causes,  say,  100  tons  compression  in  a  certain  piece,  and  if  the 

Q  O 

dead  load  is  -  of  the  live  load,  then  we  shall  have  -  of  100  == 

2  2 

150  tons  compression  in  the  same  piece  due  to  the  dead  load 
alone.  If  now  the  live  load  causes  a  maximum  tension  in  the 
same  piece  of  200  tons,  then  the  piece  must  be  made  to  resist 
both  tensile  strain  of  200  —  150  —  50  tons  and  compressive 
strain  of  150  +  100  —  250  tons.  If  a  diagonal,  the  counter  tic 
is  strained  50  tons,  while  the  maximum  strain  on  the  diagonal 


CHAP.  XII.]  METHODS    COMBINED.  181 

is  250  tons  compression.  It  is  only  necessary,  therefore,  to  find 
the  strains  due  to  each  weight  of  the  live  load.  From  the 
tabulation  we  can,  then,  by  means  of  the  ratio  of  the  dead  to 
live  load,  find  the  strains  due  to  dead  load  alone,  and  then  by 
a  comparison  of  the  two  find  the  maximum  compressive  and 
tensile  strains.  If  the  maximum  strains  due  to  live  load  are  of 
opposite  kind,  but  less  than  the  constant  strains  clue  to  dead 
load,  we  shall  need  no  counterbracing.  The  resultant  strains 
will  then  always  be  of  the  character  given  by  the  dead  load. 
If  greater,  we  must  counterbrace  accordingly.  The  process  is 
the  same  as  by  the  methods  of  calculation,  and  the  reader  may 
refer  to  Stoney — Theory  of  Strains — for  illustrations. 

12O.  The  "  Tipper,"  or  Pivot  Draw,  with  secondary  cen- 
tral Span. — We  have  said  that  a  pivot  draw  may  be  considered 
as  a  beam  continuous  over  three  supports.  In  practical  con- 
struction this  statement  needs  some  modifications  which  deserve 
special  notice.  Thus  practically  that  portion  of  the  beam  over 
the  central  support  forms  a  short  secondary  span  D  D  [Fig.  83, 
PL  22]  the  reactions  at  the  supports  D  and  D  being  always 
equal  and  of  the  same  character.  If  a  weight  acts,  say,  on  the 
first  span  A  B,  and  the  beam  itself  is  considered  without  weight, 
the  end  C  must  be  held  down,  that  is,  the  reaction  there  is  neg- 
ative. ISTow  as  the  weight  P  deflects  the  span  A  B  (Fig.  83),  it 
causes  one  secondary  support  D  to  sink,  and  the  other  to  rise 
an  equal  amount.  In  practice  D  and  D  may  be  the  extremities 
of  the  turn-table,  and  the  reactions  are  then  evidently  different 
from  those  given  by  the  formulas  of  Art.  118. 

If  in  this  case  we  take  a  as  the  distance  of  the  weight  P  from 
the  left  support  A,  the  reaction  for  load  in  A  B  will  be  given 
by  the  following  formulae : 

Where  the  ratio  -^  =  &,  a  being  the  distance  of  the  weight  P 

from  the  left  support  A  (for  load  in  the  span  A  D),  I  =  span, 
A  D  =  D  C,  and  n  1=  span  D  D,  and  where  the  constant 
(4  +  8  n  +  3  n2)  is  put  for  convenience  =  H,  then 


\  2  H  -  (10  +  15  n  +  3  n*)  &  +  (2  +  n)  &  I 
|_  J 


RA  =  —    2  H  -  (10  +  15  n  +  3  n* 
2  H 


*  See  Supplement  to  Chap.  XIII.,  Art.  6. 


182  GRAPHIC  AND   ANALYTIC  [CHAP.  XH. 

R° = 2^  r (2 + n\  ®  ~~ (2 + 3  n + 3  ^  ^  \ 

These   reactions,  it   will   be  observed,  when    added  together 
RA  4-  2  RD  +  Rc  are  equal  to  P,  as  should  be  the  case. 

By  the  application  of  these  formulae,  which  are  for  any  par- 
ticular case  by  no  means  intricate,  we  can  find  the  reactions  at 
A  and  C  as  also  at  D  or  D ;  and  then  starting,  say,  from  A,  can 
follow  the  reaction  there  through  the  frame  by  the  method  of 
Art.  114.  A  negative  reaction  indicates  that  the  support  tends 
to  rise,  and  unless  more  than  counterbalanced  by  the  positive 
reaction  due  to  uniform  load,  the  end  where  this  negative  reac- 
tion occurs  must  be  latched  down. 

121.  Supports  in  Pivot  Span  are  not  on  a  level— Reac- 
tions for  live  load,  however,  are  the  same  as  for  level  sup- 
ports.— The  three  supports  of  a  pivot  span  should  not  be  on  a 
level.  It  is  evident  that  if  this  were  the  case,  the  first  time  the 
draw  is  opened  the  two  cantilevers  deflect  and  it  would  be  diffi- 
cult to  shut  it  again.  The  centre  support  should  therefore  be 
raised  until  the  reactions  at  the  end  supports  are  zero,  that  is, 
until  they  just  bear.  The  centre  support  is  then  raised  by  an 
amount  equal  to  the  deflection  of  the  learn  when  open,  due  to 
the  dead  load.  Even  when  shut,  then,  there  are  no  reactions  at 
the  end  supports  except  when  the  moving  load  comes  on.  Now 
this  being  the  condition  of  things,  it  may  seem,  strange  to  assert 
that  these  reactions  %XQ  precisely  the  same  as  for  three  level  sup- 
ports, and  yet  such  is  the  fact.  If  the  beam,  originally  straight 
were  held  down  at  the  lower  ends  by  negative  reactions,  then 
the  reactions  would  have  to  be  investigated  for  supports  out  of 
level,  and  a  load  would  diminish  these  negative  reactions,  or 
might  even  cause  them  to  become  positive.  But  such  is  not 
the  state  of  things.  The  end  reactions  are  in  the  beginning 
zero,  and  any  load  gives,  therefore,  at  once  positive  reaction  at 
its  end  support.  This  positive  reaction  is  just  what  it  ivoulo, 
be  for  the  same  beam  over  three  level  supports. 

An  analytical  discussion  of  the  case  would  be  out  of  place 
here,  but  assuming  the  expression  to  which  such  a  discussion 
would  lead  us,  we  may  show  that  this  is  so. 

Thus,  for  a  beam  over  three  supports  A,  B,  and  C,  not  on  a 
level.  GI  being  the  distance  of  A  below  B,  and  c%  the  distance 
of  C  below  B,  the  modulus  of  elasticity  being  E  and  the  mo- 


CHAP.  XII.]  METHODS    COMBINED.  183 

inent  of  inertia  I,  we  have  for  the  moment  at  the  centre  sup- 
port B  due  to  any  number  of  weights  in  both  spans, 


a  being  always  measured  from  the  left  support. 

Now  in  this  expression  the  last  two  terms  are  precisely  the 
same  as  for  supports  on  a  level  ;  the  influence  of  the  different 
levels  is  contained  in  the  first  term  on  the  right  only.  Now  by 
the  supposition,  ct  and  c2  must  be  taken  equal  to  the  deflection 
due  to  the  dead  load,  and  the  value  of  this  term  will  therefore 
be  entirely  independent  of  the  live  load,  which  enters  only  in 
the  last  two  terms. 

A  particular  case  may  perhaps  render  this  plainer.  If  a 
girder  of  two  equal  spans  over  three  level  supports  is  uni- 
formly loaded,  the  reaction  at  an  end  support  is,  as  is  well 
known,  |ths  of  the  load  on  one  span. 

Now  let  us  take  the  girder  over  three  supports  not  on  a  level, 
and  from  our  formula  above  find  the  reaction  at  one  end  due 
to  uniform  load  when  ^  and  c2  have  the  proper  values  given 
to  them.  First  the  dead  load  p  I  over  each  span  causes  a  de- 

flection at  each  end  of  the  two  cantilevers  —  ii&s-i     This, 

then,  is  the  value  for  ^  and  <?2  m  the  formula.  Now  let  us 
take  an  additional  moving  load  of  ml  over  the  whole  beam, 
and  with  this  value  of  c1?  C2  find  the  reaction.  We  have  from 
our  formula 

4  MB  I  =  -  lp  1?  -  \  (p  +  m)  P, 
or  MB  =  -£>Z2- 

Now  we  have  by  moments, 

I2 
RAX  Z-(^  +  m)-= 

hence,  inserting  the  value  of  MB  above, 


*  Theorie  der  Trdger:  Weyranch.     Also  Supplement  to  Chap.  XIII.,  Art.  3. 
f  Supplement  to  Chap.  VII.,  Art.  13. 


184  GEAPHIC  AND  ANALYTIC          [CHAP.  XII. 

That  is,  the  reaction  at  A  is  due  to  the  moving  load  alone, 
as  evidently  should  be  the  case,  and  is,  moreover,  just  what  it 
should  ~be  for  a  girder  with  level  supports;  viz.,  \ml.  (See 
also  Appendix,  Art.  18,  Ex.  5.) 

The  raising  of  the  centre  support,  then,  will  not  affect  our 
construction  for  the  reactions  as  given  in  Figs.  81  and  82,  pro- 
vided there  are  only  three  supports. 

We  have  deemed  it  well  thus  to  call  special  attention  to  the 
considerations  of  the  last  two  articles,  both  on  account  of  their 
practical  importance  and  because  they  are  not  brought  out 
clearly,  nor  indeed,  so  far  as  we  are  aware,  ever  alluded  to  in 
any  treatise  upon  the  subject.* 

122.  Bctiiii  continuous  over  four  Level  Supports. — We 
thus  see"  that  a  draw  or  pivot  span  is  more  properly  considered 
as  a  beam  of  three  spans  instead  of  two,  of  which  the  centre 
span  is  very  small  compared  to  the  end  spans ;  it  may  be  only 
two  or  three  panels  long.  Moreover,  we  must  often  in  practice 
consider  the  beam  as  a  "  tipper,"  and  therefore  apply  the  formulae 
for  reactions  of  Art.  120.  If,  however,  by  reason  of  the  method 
of  construction,  as  often  happens,  for  instance,  by  the  under 
portion  of  the  beam  coming  in  contact  with  the  frame  below, 
this  tipping  of  D  D  (Fig.  83)  is  confined  between  certain  limits, 
beyond  which  the  supports  must  be  considered  fixed,  it  will  be 
necessary  to  find  the  reactions  as  for  a  beam  over  four  fixed 
supports,  and  determine  the  corresponding  strains  in  this  case- 
also. 

Comparing,  then,  the  strains  obtained  each  way,  we  take  only 
the  maximum  strains  from  each. 

The  formulae  for  the  reactions  at  \h.Q  fixed  supports  A  B  C  D 
are  as  follows  (PI.  22,  Fig.  84) : 

1st.  Load  P  in  left  end  span  A  B  at  a  distance  a  from  left: 
support  A,  the  end  spans  being  n  I  and  the  centre  span  B  C=L 

We  put  Jc  =  —~  and  H  =  3  +  8  n  +  4  n\     Then 
n  L 

RA=g|  H-(H  +  2^-h2^)&-f(2rc+2rc,2)F 


*  Clemens  Herschel,  in  his  treatise  upon  "  Continuous,  Revolving  Draw- 
bridges" (Little,  Brown  &  Co.,  Boston,  1875),  notices  this  fact  for  the  first 
time. 


CHAP.  XII.]  METHODS   COMBINED.  185 

RB=j||   (3  +  10^+9™2+2^3)&-(27i  +  5^2+27i3)&3 
Rcr=-|    -(n+3n*  +  2n*)fc+(n  +  Sn*+%n*)& 

RD=H|  njc~n^ 

These  reactions  add  up,  as  they  should,  equal  to  P. 

In  practical  cases  of  pivot  spans,  we  have  only  to  consider  the 
the  outer  spans ;  as  a  load  in  the  middle  span  B  C  =  I  rests 
directly  upon  the  turn-table.  The  above  formulae  are  then  all 
we  need.  For  a  load  in  the  right  end'  span  the  same  formulae 
hold  good,  only  remembering  to  put  now  RD  in  place  of  R^ 
Rc  in  place  of  RB,  RB  in  place  of  Rc,  and  RA  in  place  of  RD. 

If,  however,  neglecting  the  particular  case  of  pivot  spans,  we 
suppose  the  middle  span  B  C  =  I  loaded,  we  have — a  being 

now  the  distance  of  P  from    B,  and  &  being  now  -j  instead 

C 

of  — ^,  as  above,  H  remaining  the  same. 
2d.  Load  in  B  C. 


1 

LH     3    4n-M-4n*   *- 6    15       62         H     1 

[si 
J 

r    ^  '  _  ^      ^    i 
L  J 


These  reactions  should  also  add  up  to  P,  as  is  the  case.  The 
number  n  may  be  taken  at  pleasure,  so  that  the  end  spans  may 
be  as  much  larger  or  less  than  the  centre  spans  as  is  desired. 

H,  P  and  the  quantities  in  the  parentheses,  it  will  be  observed, 
are  for  any  given  case,  constants  which  may  be  determined  and 
inserted  once  for  all. 

"We  have,  then,  only  to  insert  the  values  of  7c  for  different 
positions  of  the  load  P.  Thus  the  equations  for  any  particular 
'•ase'are  very  simple  and  easy  of  application. 


186  GRAPHIC   AND   ANALYTIC  [CHAP.  XII. 

123.  Construction.  —  We  may,  if  desired,  apply  our  method 
of  construction  to  the  determination  of  the  reactions.  Thus 
from  the  above  reactions  we  may  easily  determine  general  ex- 
pressions for  the  inflection  points.  For  the  case  of  a  load  in 
C  D  =  n  I  (PI.  22,  Fig.  85),  we  have,  when  i  is  the  distance  of 
the  inflection  point  from  C, 

-RD  x  (nl-i)  +P(a-i)  =  0; 

P  a  -  RD  n  I 

whence  ^  =  —=  -  =  —  . 

P  —  KJJ 

For  the  inflection  point  distant  i  from  B  in  the  unloaded 
span, 


hence 


For  the  second  case  of  load  in  B  C  =  I,  we  have  for  the  in- 
flection point  between  B  and  P 

—  RA  (n  l+i)  +  Rji  i  =  0,  or 


i=K±nl_ 


For  the  point  between  P  and  C 

—  RD  (n  I + i)  +  Rc  i  =  0,  or 


The  insertion  of  the  proper  values  of  the  reactions  for  each 
case,  as  given  above,  will  easily  give  general  expressions  for  the 
inflection  points,  which  the  reader  may,  if  desired,  deduce  for 
himself. 

Our  construction  is,  then,  as  follows  [PL  22,  Fig.  84]  : 

\st  Case.     Load  in  C  D.  -. 

Having  found  %,  draw  a  line  at  any  inclination,  as  ^  d  through 
&1?  intersecting  P  at  d,  and  the  vertical  through  C  at  c^  Then 
lay  off  B  i  and  draw  d  D,  c±  b  and  ~b  A. 

Make  d  c  =  P  by  scale,  and  c  D  drawn  parallel  to  ^  d  then 
gives  the  pole  distance  H.  The  ordinates,  then,  to  the  broken 
line  A  b  c:  d  D  taken  to  scale  of  distance,  multiplied  by  H  to 
scale  of  force,  give  the  moments  at  every  point.  Moreover,  H  d 
is  the  reaction  at  D.  Draw  D  b  parallel  to  c±  5,  then  c  b  is  the 
reaction  at  C.  In  like  manner  a  b  is  the  reaction  at  B,  and  H  a 
the  reaction  at  A.  The  moment  at  C,  and  reactions  at  C  and 


CHAP.  XII.]  METHODS    COMBINED.  187 

D,  are  positive.  Reaction  and  moment  at  B  negative  /  reaction 
at  A  positive ;  as  a  little  consideration  of  what  the  curve  of  the 
deflected  beam  must  be,  will  show.  The  shear  at  C  is,  there- 
fore, +  H  a  —  a  b  +  b  c  —  +  H  c.  The  shear  at  B  is  — H  b  or 
4-  H  a— a  b,  and  so  on.  The  shear  being  always  given  by  the 
segment  between  H  and  lines  parallel  to  A  b,  b  cl}  c{  d  and  d  D. 

2d  Case.— PL  22,  Fig.  85.— Load  in  B  C. 

Having  found  the  distance  B  i±  from  the  equation  for  this  dis- 
tance of  the  point  of  inflection  above,  we  lay  off  B  c\  =  B  -&t 
and  thus  draw  c±  E  at  an  angle  of  45°.  Finding  then  the  value 
of  C  i2  from  its  equation  above,  we  can  draw  E  c%  and  then 
c2  D  and  Ci  A.  The  construction  is  then  the  same  as  before. 
Thus  H  is  the  pole  distance,  H  c  the  negative  reaction  at  D, 
H  b  the  negative  reaction  at  A,  and  cl  b,  c  E  the  positive  re- 
actions at  B  and  C.  The  shear  at  B  is  H  cly  etc.  Thus  the  outer 
forces  are  completely  known  for  a  weight  at  any  point.  It  will, 
however,  in  general,  in  practice,  be  found  more  satisfactory  to 
use  the  formulae  for  the  reactions  which  we  have  given  than  to 
find  these  reactions  by  the  above  construction. 

We  shall  now  illustrate  the  preceding  principles  by  an  exam- 
ple taken  from  actual  practice. 

124.  »raw  Span— Example.— In  PL  22,  Fig.  86,  we  have 
given  to  a  scale  of  20  ft.  to  an  inch  the  elevation  of  one  of  the 
trusses  of  the  pivot  draw  over  the  Quinnipiac  Eiver  at  Fair 
Haven,  Conn.* 

Length  of  span  A  B  =  89.88  ft.  B  C  =  21.666  ft,  divided 
into  seven  panels  of  12.84  ft.  and  two  of  10.833  ft.  respectively. 

Height  at  B  and  C,  16  ft. ;  at  A  and  D,  12.1  ft.  Diagonal 
bracing  as  shown  in  Fig.  Line  load  9  tons  per  panel. 


21  ( 
In  this  case  n  =  — '—  -  or  n  =  0.24106,  hence  the  equations 

of  Art.  120  become 

A  =  P  (l-1.1298  ?+0.1836  ^ 

B 


=  C  =  P  /0.6S36  2-0.1836 


*  Designed  and  erected  by  Clemens  Herschel,  C.E.,  and  probably  the  only 
structure  of  the  kind  in  this  country  for  which  the  strains  have  been  accu- 
rately and  thoroughly  determined.  For  the  above  data  I  am  indebted  to  M. 
Merriman,  assistant  engineer  in  charge. 


188  GRAPHIC  AND  ANALYTIC         *  [CHAP.  XII. 

D  =  -P  fo.2374  —0.1836  -V 

\  \  I  1?  I 

Now  -'  is  -,  -,  -,  -ths,  etc.,  according  to   the  position   of    the 

L        7    i    7    * 

weight  at  lst%  2d,  3d  apex  from  end.    So  also  —  is  — r,  — ,  — — 

/  Otto     343     343, 

etc.      The   above   equations  for   the   reactions,  then,  may  be 
written  A  =  P  (1-0.1614  £  +  0.000535  £3), 

B  =  C  =  P  (0.09766  £-0.000535  £3), 
D  =  -P  (0.03391  £-0.000535  £3), 

where  £  has  the  values  1,  2,  3,  4,  etc.,  for  P!,  P2,  P8,  P4. 

Thus,  if  we  wish  the  reactions  due  to  a  weight  P4  of  9  tons  at 
the  fourth  apex,  as  shown  in  Fig.,  we  have  only  to  make  P  —  9 
and  £  —  4,  and  we  find  at  once  A  =  3.498  tons,  B  =  C  = 
3.207  tons,  D  =  —0.912  tons.  The  sum  of  all  these  reactions 
exactly  equals  P,  as  should  be. 

The  middle  supports  are  supposed  raised  by  an  amount  equal 
to  the  end  deflections  of  the  open  draw,  therefore  the  strains 
due  to  dead  load  are  easily  found,  as  in  the  "  braced  semi-arch," 
Art.  9. 

The  reactions  due  to  live  load,  according  to  Art.  121,  will  not 
be  affected  by  this  raising  of  the  supports. 

To  find  the  strains  due  to  P4,  we  draw  the  force  line  F2  F 
[Fig.  86  (a)]  by  laying  off  P4  —  9  tons  down  from  F  to  Fl3  then 
F!  E2  downwards  equal  to  the  negative  reaction  at  L ,  viz., 
—0.912  tons.  Then  from  E2  lay  off  upwards  E2  Et  =  to  positive 
reaction  at  C  — +3.207  tons.  Then  ^  E  =  reaction  at  B  = 
+  3.207  tons,  and  finally  E  F  equal  to  reaction  at  A  =  +3.498 
tons,  which  should  bring  us  back  exactly  to  point  of  beginning 
F,  since  the  reactions  and  the  weight  P  must  be  in  equilibrium. 
\JN~ote. — When  we  wish  to  begin  at  the  left  end  of  the  frame, 
it  is  best,  as  in  this  case,  to  lay  off  the  reactions  in  order,  com- 
mencing at  the  right.']  We  have  taken  the  scale  of  force  4 
tons  per  inch. 

The  weight  P4  acts  upon  the  triangulation  drawn  full  in  the 
figure.  Using  now  the  notation  of  Art.  114,  and  representing 
all  the  space  above  the  truss  by  E,  all  lelow  by  F,  we  have  at 
A  the  reaction  E  F  [Fig.  86  (a)]  in  equilibrium  with  E  1  and 
F  1,  and  drawing  parallels  to  these  lines  from  E  and  F,  we  find 
the  strain  in  F  1  =  3. 54  tons  tension,  and  E  1  =  5.1  compression. 


CHAP.  XII.]  METHODS    COMBINED.  ISO 

So  we  go  through  the  truss  and  find  the  strains  in  every 
piece.  Heavy  lines  in  the  strain  diagram  denote  compression. 
We  see  at  once  that  for  this  position  of  the  weight,  all  the 
upper  flanges  in  span  A  B  are  compressed,  the  last  lower  flange 
P  7  is  also  compressed,  and  all  the  other  lower  flanges  are  in 
tension.  At  the  point  of  application  of  the  weight  P4,  the  two 
diagonals  3  4  and  4  5  are  in  tension,  and  either  side  they  alter- 
nate in  strain  as  far  as  C  or  diagonal  8  9.  Diagonals  8  9  and  9  10 
are  both  tension,  and  then  the  strains  alternate  to  support  D. 
All  the  upper  flanges  of  the  right  half  are  tension  and  increase 
towards  the  middle.  All  the  lower  are  compression  and  like- 
wise increase  towards  the  middle. 

If  we  go  through  the  whole  truss  from  A  to  D,  the  last  diago- 
nal 15,  16  should  evidently  pass  exactly  through  E2,  thus  check- 
ing the  accuracy  of  the  construction.  The  diagonal  6  7  crosses 
the  force  line,  thus  causing  the  strain  in  the  lower  flange  to 
change  from  tension  in  Ft  5  to  compression  in  F!  7.  The  point 
of  inflection,  therefore,  falls  to  the  right  of  diagonal  5  6. 

The  reaction  at  B  diminishes  greatly  the  strain  which  would 
otherwise  take  effect  in  7  8  and  E  8 ;  while  the  reaction  at  O 
reverses  the  strain  which  would  otherwise  take  effect  in  9  10 
and  diminishes  E  10.  We  recommend  the  reader  to  follow 
through  carefully  the  strain  diagram,  Fig.  86  (a). 

A  series  of  figures  similar  to  Fig.  86  (a)  (in  the  present  case 
seven  separate  figures)  will  give  completely  the  strains  due  to 
the  rolling  load.  A  table  may  then  be  drawn  up  containing 
the  strains  due  to  dead  load,  and  the  maximum  strains  due  to 
live  load  in  every  piece,  and  the  total  maximum  tension  and 
compression  in  every  piece  may  then  be  found.  [Compare 
Art.  12,  Fig.  7.] 

For  the  supports  fixed,  instead  of  B  and  C  tipping,  the  pro- 
cess is  precisely  similar,  except  that  we  have  to  make  use  of 
the  formulae  of  Art.  12i4.  The  reaction  at  A  will  then  be 
somewhat  less  than  in  the  present  case ;  the  inflection  point  is 
therefore  found  further  from  the  right  support  B ;  it  may  be 
even  to  the  left  of  diagonal  5  6,  in  which  case  (see  Fig.  86,  a) 
we  should  have  tension  in  upper  flange  E  6.  The  reaction  at 
B  would  then  be  still  positive,  but  greater  than  E  E1?  while  C 
would  be  negative  and  no  longer  equal  to  B,  and  D  would  be 
positive.  We  should  thus  have  7  8  tension  and  E  8  tension  ; 
F  7,  as  before,  compression,  8  9  compression,  and  9  10  com- 


190  GRAPHIC    AXD   ANALYTIC  [CHAP.  XII. 

pression,  and  E  10  compression ;  while  F  9  would  be  tension. 
From  9  10  to  the  right  the  diagonals  would  alternate  in  strain, 
the  compressed  upper  flanges,  as  also  the  tensile  lower  flanges, 
would  diminish  towards  D,  and  the  last  diagonal  should  pass 
exactly  through  new  position  of  F2,  thus  closing  the  strain 
diagram  and  checking  the  work.  The  reader  will  do  well  to 
construct  the  diagram. 

The  strains  should  be  found  for  both  cases,  and  the  maximum 
strains  taken  from  each,  which,  compared  with  the  permanent 
strains  due  to  the  dead  load,  will  give  the  total  maximum 
strains. 

"We  have  taken  for  convenience  of  size  too  small  a  scale  for 
the  frame  to  ensure  good  results.  With  a  large  and  accurately 
constructed  ymw e  diagram,  dealing  as  we  do  with  only  single 
weights,  and  consequently  small  strains,  the  above  force  scale 
of  4  tons  per  inch  would  give  very  accurate  results. 

If  the  strains  due  to  uniform  load  (no  end  reactions)  are 
found  by  addition  of  the  strains  for  each  apex  load  diagramed 
separately,  the  same  scale  may  be  employed ;  but  if  all  the 
loads  are  taken  as  acting  together  (Fig.  5,  b\  a  smaller  scale 
for  strains  will  have  to  be  adopted,  as  the  force  line  will  other- 
wise be  too  long.  [See  Art.  16  of  Appendix  for  the  method  of 
calculation.] 

125.  Method  of  passing  direct  from  one  Spun  to  next. 

— By  inspection  of  Fig.  86  we  see  that  we  might  find  the  strains 
in  the  intermediate  span  B  C  without  first  going  through  the 
whole  of  A  B  or  C  D.  Thus,  if  we  knew  the  moment  at  B, 
this  moment^  divided  by  depth  of  truss  at  B,  would  give  the 
strain  in  flange  F  7  for  the  system  of  triangulation  indicated 
by  the  full  lines.  If  then  we  knew  also  the  shear  at  B  = 
P  —  A  —  B  =  E!  F!  (Fig.  86,  a),  we  could  at  once  lay  off  Fl  7 
and  E!  Ft  (Fig.  86,  a\  and  then  proceed  to  find  E  8  and  7  8, 
just  as  before.  In  the  same  way  the  moment  at  C,  divided  by 
height  of  truss  at  C,  would  give  us  strain  in  F  9,  and  with 
shear  at  C  =  P  -  A  -  B  -  C  =  E2  Fl  =  D,  we  could  find  E  10 
and  9  10,  as  before.  As  we  know  already,  a  load  anywhere 
upon  a  beam  causes  positive  moments  at  a  fixed  end — i.e., 
makes  upper  flange  over  support  tension  and  lower  flange  com- 
pression. But  as  we  see  from  the  last  case,  owing  to  the  tri- 
angulation, the  last  upper  flange  may  also  be  compression  (see 
E  6  in  Fig.  86)  if  the  inflection  point  lies  between  diagonal  5  6 


CHAP.  XH.J  METHODS    COMBINED.  191 

and  the  support.  The  known  moment  gives,  then,  the  charac- 
ter of  the 'strain  only  for  that  flange  which  does  not  meet  a 
diagonal  at  the  support.  The  moment  at  B,  therefore,  being 
positive,  gives  us  compression  here  in  lower  flange,  because,  for 
the  system  of  triangulation  corresponding  to  the  weight,  that 
flange  does  not  meet  a  diagonal  at  B.  For  a  weight  upon  the 
other  system  of  triangulation  (dotted  in  Fig.),  the  same  moment 
would  give  us  the  tension  in  E  7.  The  construction  assumes 
equilibrium  between  F  7,  7  8,  and  E  8,  and  the  shear  at  B ; 
that  is,  between  the  pieces  cut  by  an  ideal  section  to  the  right 
of  B  through  the  truss  and  the  shear  at  that  section.  That  this 
is  so  is  shown  by  the  strain  diagram,  since  there  we  see  that  the 
strains  in  these  pieces  form  a  closed  polygon  with  the  shear  at 
B  =  E!  Fx.  This  must  evidently  be  so  if  these  are  the  only 
pieces  cut  by  such  a  section,  since  then  the  horizontal  com- 
ponents of  the  strains  in  these  pieces  must  balance,  and  the 
resultant  vertical  component  must  be  equal  and  opposite  to  the 
shear. 

It  is  important  to  know  which  side  of  Et  Pj_  to  lay  off  F!  7, 
since,  if  we  had  laid  it  off  in  this  case  to  the  right,  we  would 
have  obtained  a  very  different  value  for  Et  8.  For  this  pur- 
pose we  have  only  to  suppose  the  strain  in  the  flange  (either 
upper  or  lower,  as  the  case  may  be)  to  be  applied  at  the  point 
of  junction  or  apex  of  the  other  two  pieces,  and  then  lay  it  off 
in  the  direction  with  reference  to  that  apex  corresponding  to 
the  known  character  of  its  strain.  The  direction  of  the  shear 
is  always  known  from  the  reactions. 

Thus  in  our  Fig.  the  shear  between  B  and  C  acts  down  from 
E!  to  Fl5  because  P4,  which  also  acts  down,  is  greater  than  the 
sum  of  the  upward  reactions  at  A  and  B.  The  strain  in  F  7 
is  also  known  to  be  compressive,  and  therefore,  in  following 
round  the  strain  polygon  commencing  from  Ex  to  F1?  it  must 
act  towards  apex  at  7.  We  must,  therefore,  lay  it  off  to  the 
left  of  E!  Fj.  In  similar  manner,  for  the  other  triangulation, 
the  strain  in  flange  E  7  is,  in  span  B  C,  in  equilibrium  with  7  8 
(dotted  diagonal)  and  P  8  and  shear  Et  F1?  and  is,  moreover, 
known  to  be  tension.  Consider  it  acting  then  at  B ;  and  then, 
since  it  is  tension,  we  go  round  the  polygon  from  Et  to  Pt,  and 
then  to  the  right  of  Ex  F1?  or  away  from  B,  the  point  at  which 
it  is  supposed  to  act. 

Now  for  the  case  of  the  "  tipper : "  the  reaction  at  D,  and 


192  GKA.PHIC   AND   ANALYTIC  [CHAP.  XII. 

therefore  the  moment  at  C,  is  also  positive.  The  lower  flange 
F  9  is  therefore  compression,  or  for  the  dotted  system  of  trian- 
gulation  E  9  is  tension.  The  shear  to  the  left  of  C,  Et  F!  acts 
down,  since  —  P  +  A  +  B  =  —  ^  Flt  Consider  F  9  acting  at 
apex  9,  and  then,  since  it  is  compression,  it  must  act  towards  9 
(from  right  to  left),  and  passing  down  then  from  Ex  to  F1?  we 
must  lay  off  F  9  to  the  left  of  Ex  F^  For  similar  reasons,  for 
the  other  system,  E  9  must  be  laid  off  to  the  right. 

For  fixed  supports  B  and  C,  the  moments  alternate  from  B, 
and  the  moment  at  C  is  therefore  negative — that  is,  gives  com- 
pression above  and  tension  below.  Flange  F  9,  for  the  system 
of  triangulation  of  P,  would  then  be  tension  instead  of  com- 
pression, as  above ;  P  will,  however,  still  be  greater  than  A  +  B, 
and  hence  the  shear  is  to  be  laid  off  down,  and  F  9  must  be 
laid  off  to  the  right. 

If,  then,  it  were  required  to  find  the  strains  in  the  span  B  C, 
preceded  and  followed,  it  may  be,  by  many  others,  it  is  suffi- 
cient to  know  the  moment  and  shear  at  one  support.  We  can 
then  commence  and  continue  the  strain  diagram,  without  being 
obliged  to  go  off  to  a  distant  free  end  and  trace  all  the  strains 
through  till  we  arrive  at  the  span  in  question. 

126.  Method  of  procedure  for  any  number  of  Spans. 
Let  us  take,  then,  any  number  of  spans,  say  seven  [PL  22,  Fig. 
87],  and  let  it  be  required  to  find  the  maximum  strains  in  the 
span  D  E.  It  is  not,  as  we  have  just  seen,  necessary  to  com- 
mence at  the  extreme  end  A  or  H,  and  follow  the  reaction  there 
through,  from  span  to  span,  till  we  arrive  at  D.  As  we  have 
seen  from  the  preceding  Art.,  we  may  start  directly  from  D, 
provided  we  know  the  moment  and  shear  there.  Now,  since  a 
load  in  any  span  causes  positive  moments  and  reactions  at  the 
two  ends  of  that  span,  and  since  either  way  from  these  ends  the 
moments  and  shear  at  the  other  supports  alternate  in  character 
[Art.  102],  any  and  all  loads  in  A  B  cause  positive  moments 
and  reactions  at  D.  So  also  for  loads  in  C  D  and  in  F  G. 
Loads  in  B  C,  E  F  and  G  H,  on  the  other  hand,  cause  negative 
moments  and  reactions  at  D.  [See  Fig.  87.] 

To  find  the  maximum  positive  moment  and  shear  at  D  due 
to  the  other  spans,  we  must  then  suppose  the  method  of 
loading  shown  in  Fig.  87  (a).  For  the  maximum  negative 
moment  and  shear  at  D,  we  have  the  system  of  loading  shown 
in  Fig.  87  (b). 


CHAP.  XII.]  METHODS    COMBINED.  193 


these  two  moments  and  shears  being  once  known,  we 
can  find  by  diagram  and  tabulate  the  respective  strains  in  every 
piece  of  the  span  D  E.  Thus  dividing  the  moment  at  D  for 
either  case  by  the  height  of  truss,  we  have  at  once  the  strain  in 
either  upper  or  lower  flange  at  D  depending  upon  the  system 
of  triangulation  as  explained  in  Art.  125.  With  this  strain  and 
the  shear  at  D  properly  laid  off  to  scale,  we  can  commence  the. 
strain  diagram  precisely  as  though  we  had  traced  all  the  loads 
through  from  the  extreme  end  A  or  H  to  D  or  E. 

We  must  next  find  and  tabulate  the  strains  in  D  E  due 
to  each  apex  load  in  the  span  itself,  and  for  this  we  must 
know  to  begin  with  the  moments  and  shears  for  each  separate 
load. 

[Note.  —  Distinguish  carefully  between  shear  and  reaction  at 
a  support.  The  shear  at  D,  or  at  a  point  just  to  right  of  D,  is 
the  algebraic  sum  of  all  the  reactions  and  weights  between  that 
point  and  A.  See  also  Fig.  84  (Art.  123),  where  the  reaction 
at  B  is  —  l>  a,  but  the  shear  atB  is—  b  a  +  H.  a  =  —  H  b. 
So  also  the  reaction  at  C  is  -f  ~b  c,  but  the  shear  at  C  is 
+  I  c  —  ba  +  Ha  —  HC,  etc.] 

Conceiving  n-ow  that  we  have  found  and  tabulated  the  strains 
due  to  the  first  and  second  systems  of  loading  as  shown  in  Fig. 
87,  and  also  the  strains  for  each  load  P  in  D  E,  the  sum  of  these 
strains  will  give  the  strains  due  to  live  load  o-ver  the  whole 
length  of  girder,  and  taking  the  proper  proportion  of  these,  we 
shall  have  the  strains  due  to  the  dead  load.  Combining  then 
these  strains  with  those  first  found,  we  can  easily  find  the  total 
maximum  strains  which  can  ever  occur  in  D  E. 

127.  Example.  —  Let  us  take,  as  an  illustration  of  the  pre- 
ceding, the  girder  shown  in  Fig.  87,  of  seven  equal  spans,  and 
seek  the  maximum  strains  which  can  ever  occur  in  the  middle 
span  D  E.  Let  Fig.  88,  PI.  23,  represent  the  span  D  E  —  length 
80  feet,  divided  into  4  panels  ;  and  let  the  live  load  at  each 
apex  be  40  tons,*  the  uniform  load  being  half  as  much,  or  20 
tons  per  apex.  Height  of  truss  =  10  feet. 

Now  the  quantities  which  for  the  present  we  must  suppose 
known  or  already  found  are  as  follows  : 


*  A  very  great  load  :  half  the  resulting  strains  would  give  more  nearly  the 
strains  in  a  single  truss. 
13 


194  GRAPHIC   AND   ANALYTIC  [CHAP.  XII. 

Positive  moment  at  D  (1st  system  of  loading), 

as  shown  by  Fig.  87  (a)      ....—  +  788.2  feet  tons. 

Corresponding  shear  at  D =  -1-  14.63  tons. 

Negative  moment  at  D 

(2d  system  of  loading) =  -  382.54. 

Corresponding  shear ~  —  14.63. 

Also  for  the  loads  in  D  E : 

For  the  first  load  P1?  moment  —  -f  158.92,  shear  =  +  36.17. 
P,  «  =  +  271.96  «  =  +  25.88. 
P3  "  =  +  203.36  «  =  +  14.16. 
P4  "  =  +  62.88  "  =  +  3.82. 

In  Fig.  88  we  have  found  by  diagram  the  strains  due  to  P3. 
[For  notation,  see  Art.  114.] 

We  l&y  off  to  scale  the  shear  14.16  upwards,  since  it  is  posi- 
tive, and  then,  since  the  moment  203.36  at  D  is  positive,  and 
hence  the  strain  in  A  a  must  be  tension,  we  lay  off  A  a  = 

?H^-  =  20.3  tons  to  the  right  of  B  A  (Art.  125).      With  B  A 

and  A  a  thus  given,  we  can  rapidly  and  accurately  find  all  the 
other  strains.  Thus  from  our  diagram  we  have,  representing 
tension  by  minus  and  compression  by  plus, 

A  a  =  —  20.3     A  c  —  +  8.0    A  e  =  +  36.4      A  g  =  +  24.4 
A  ]c  =  -  27.2 

B6=  +  6.0    B<Z=:-22  B/=-50.8    B  A  =  +  1.2  tons ; 

and  for  the  diagonals : 

a  &=+19.6   be  =-19.6  cd  =  +  19.6    de=-l9.6 

ef  =  +  19.6  fg  =4-  36.4  g  h  =-36.4    h  Jc  =  +  36.4  tons. 

Heavy  lines  in  the  diagram  represent  compression. 

In  a  manner  precisely  similar  we  can  find  the  strains  due  to 
the  other  weights,  as  also  to  the  two  systems  of  loading  shown 
in  Fig.  87.  Suppose  all  these  strains  thus  found.  Then  the 
method  of  tabulation  is  as  follows : 


CHAP.  XII.] 

123 


METHODS 
56 


10 


11 


195 

12 


DIAGONALS.  FLANGES. 

'l 

Live  Load  in  D  E  - 

Interior  Loading 
Maximum 
Strains. 

Exterior  Load- 
ing. 

Dead 
Load 
=»# 
Live. 

Total 
Maximum 
Strains. 

P: 

*  2 

p3 

?4 

Tens. 

Comp. 

+ 

1st 
Case. 

2d 
Case. 

Tens. 

Comp. 

+ 

Aa 

-  15.6 

-  27.2 

_  20.4 

-    6.4 

-  69.6 

-78.8 

+  38.2 

-55.1 

-  203.5 

+  63.6 

Ac 

+  16.4 

+  24.4 

+    8.0 

+    1.2 

+  50.0 

-  49.5 

+    8.9 

+    4.7 

-44.8 

Ae 

4-    8.8 

+  36.4 

+  36.4 

+    8.8 

.... 

+  90.4 

-20.3 

-  20.3 

+  24.9 

-  15.7 

+  115.3 

Afir 
A* 

+     1.2 

+    8.0 

+  24.4 

+  16.4 

+  50.0 

+    8.9 

-  49.5 

+    4.7 

-44.8 

+  63.6 

-    6.4 

-  20.4 

-27.2 

-  15.6 

-  69.6 

+  38.2 

-  78.8 

-  55.1 

_  203.5 

B6 

-20.4 

+    1.2 

+    6.0 

+    2.8 

-20.4 

+  10.0 

+  64.2 

-  23.6 

+  15.1 

-28.9 

+  89.3 

B<1 

-12.8 

-50.8 

-22.0 

-    5.2 

-90.8 

+  34.9 

+    5.6 

-25.1 

-115.9 

+  15.4 

B/ 

-    5.2 

-  22.0 

-  50.8 

-12.8 

-90.8 

+    5.6 

+  34.9 

-  25.1 

-  115.9 

+  15.4 

B/, 

+    2.8 

+    6.0 

+    1.2 

-  20.4 

-20.4 

+  10.0 

-  23.6 

+  64".  2 

+  15.1 

-28.9 

+  89.3 

a  6 

+  51.2 

+  36.4 

+  19.6 

+    5.2 



+  112.4 

+  20.7 

-  20.7 

+  56.2 

+  189.3 

fo  c 

+    5.2 

-  36.4 

-  19.6 

-    5.2 

-61.2 

+    5.2 

-20.7 

+  20.7 

-  28.0 

_  109.9 

<:  d 

-    5.2 

+  36.4 

+  19.6 

+    5.2 

-    5.2 

+  61.2 

+  20.7 

-  20.7 

+  28.0 

+  109.9 

d  e 

+    5.2 

+  19.6 

-  19.6 

-    5.2 

-24.8 

+  24.8 

-20.7 

+  20.7 

0.0 

-  45.5 

+  45.5 

ef 

—    5.2 

-19.6 

+  19.6 

+    5.2 

-24.8 

+  24.8 

+  20.7 

-  20.7 

0.0 

-  45.5 

+  45.5 

f(J 

+    5.2 

+  19.6 

+  86.4 

-    5.2 

-    5.2 

+  61.2 

-20.7 

+  20.7 

+  28.0 

.... 

+  109.9 

gh 

-    5.2 

-19.6 

-36.4 

+    5.2 

-  61.2 

+    5.2 

+  20.7 

-  20.7 

-28.0 

-  109.9 

hk 

-    5.2 

+  19.6 

+  36.4 

+  51.2 

+  112.4 

-20.7 

+  20.7 

+  56.2 

+  189.3 

Having  found  and  tabulated  the  strains  clue  to  each  weight 
as  shown  by  the  first  five  colums  of  the  table,  add  all  the  ten- 
sions and  compressions  for  each  piece  and  place  the  results  in 
columns  6  and  7.*  ^Ve  thus  have  the  maximum  strains  of  each 
kind  which  can  be  caused  by  the  weights  in  span  D  E  alone. 
In  the  next  two  columns,  8  and  9,  place  the  strains  due  to  the 
two  cases  of  loading  of  Fig.  87.  Now  if  the  uniform  or  dead 
load  is  taken  at  one  half  the  live,  we  have  simply  to  take  the 
algebraic  sum  of  the  strains  in  columns  6,  7,  8,  and  9  horizon- 
tally, and  divide  by  2.  We  thus  find  column  10.  Finally, 
from  columns  6,  7,  8,  9  and  10,  we  can  find  the  total  maximum 

*  A  more  convenient  form  of  table,  perhaps,  is  obtained  by  putting  the 
weights  Pj_4  in  the  vertical  left-hand  column,  and  the  pieces  A  a,  Ac, 
etc.,  in  the  top  horizontal  line.  The  numbers  are  thus  more  conveniently 
placed  for  add  tion. 


196  GKAPHIC    AND   ANALYTIC  [CHAP.  XII. 

strains  as  given  in  the  last  two  columns.  Thus  take  the  piece 
A  c.  In  this  piece  there  is  a  constant  compression  due  to  dead 
load  of  4.7  tons.  The  second  system  of  loading  adds  to  this 
for  live  load  8.9  tons,  and  the  loading  of  D  E  itself  causes  50 
tons  compression.  Since  all  three  cases  may  exist  together,  we 
have  4.7  + S. 9 +  50  —  63.6  tons  compression.  Again  the  first 
system  of  loading,  which  may  act  also  alone,  causes  49.5  tons 
tension  in  A  c.  Diminishing  this  by  the  4.7  tons  compression 
dne  to  dead  load  already  existing,  we  have  —49.5+4.7  =—44.8 
tons  tension.  These  two  strains  are  the  greatest  which  can  ever 
occur  in  A  c. 

For  A  a  we  see  that  the  greatest  compression  is  due  to  the 
second  system  of  loading  acting  alone,  but  this  compression  = 
'38.2  tons,  is  less  than  the  tension  always  existing  in  A  a  from 
the  dead  load,  which  is  55.1  tons.  Compression,  then,  can  never 
occur  in  A  a.  So  also  for  A  Jc.  In  like  manner  diagonals  a  £, 
b  c,  c  d,  y  <7,  g  A,  and  h  k  do  not  need  counterbracing,  as  the 
constant  dead  load  strain  overbalances  any  strain  of  opposite 
character  which  can  ever  come  upon  them. 

As  we  have  in  the  present  example  taken  a  middle  span, 
observe  that  the  strains  of  Pt  and  P4,  P3  and  P2  are  similar.  Thus, 
strain  in  A  a  due  to  P4  is  the  same  as  in  A  k  due  to  Pl5  and  so  on . 

1£§.  Method  of  Moments. — We  can  very  easily  check  our 
results  by  the  method  of  moments  of  Art.  14. 

Thus,  for  the  first  system  of  loading,  we  have  the  moment  at 
D  =  +788.2  ft.  tons  and  the  shear  =14.63.  For  any  upper 
flange,  then,  as  A  g,  since  the  positive  moment  causes  tension 
in  upper  fiange,  and  positive  shear  causes  compression,  we  have, 
taking  apex  g  as  centre  of  moments,  —  788.2  +  14.63  x  60  = 
-  788.2  +  877.8  =  +  89.6.  Dividing  this  by  depth  of  truss 
=  10  ft.,  we  have  8.96  tons  compression  in  A  g. 

For  the  strain  in  By  due  to  the  weight  P2,  since  the  moment 
at  D  for  this  case  is  +271.96  and  shear  =+25.88,  and  the 
positive  moment  and  shear  cause  compression  and  tension  re- 
spectively in  lower  flanges,  while  the  weight  P2=40  causes 
compression  in  By  we  have,  taking  f  as  centre  of  moments, 
+  271.96-25.88x50+40x20  =  -222.0:1:,  rind  dividing  this  by 
10,  we  obtain  22.2  tension  in  By,  and  so  on. 

For  the  diagonals,  the  shear  at  any  apex,  multiplied  by  the 
secant  of  the  angle  with  the  vertical,  gives  at  once  the  strain. 
In  this  case  the  angle  is  45°;  therefore  the  secant  is  1.414. 


CHAP.  XII.]  METHODS    COMBINED.  197 

Hence,  for  strain  in  c  d  due  to  P2,  we  have 
25.88x1.414:=  +36.6  tons. 

The  calculation,  then,  of  the  strains  in  any  span  of  a  continu- 
ous girder,  as  also  the  diagraming  of  these  strains,  is  simple  and 
easy,  and  offers  no  more  difficulty  than  in  the  case  of  a  simple 
truss,  provided  we  know  or  can  find  the  moments  and  shearing 
forces  at  the  supports  for  the  various  cases  of  loading.  The 
method  of  finding  these  necessary  quantities  will  form  the  sub- 
ject of  the  next  chapter. 

The  reader  will  do  well  to  compare  the  strains  in  the  above 
table  with  those  for  the  same  simple  girder  similarly  loaded.* 

It  will  be  found  that  there  is  a  saving  of  material  in  the 
flanges  of  about  eleven  per  cent,  over  the  corresponding  simple 
girder.  Some  of  the  flanges  are,  to  be  sure,  subjected  to  both 
tensile  and  compressive  strains,  instead  of  being  always  of  the 
same  character  ;  but  in  wrought-iron  girders  this  is  of  little  con- 
sequence. 

It  is  worthy  of  remark  that  a  slight  relative  difference  of  level 
of  the  supports  of  a  continuous  girder  may  cause  very  great 
changes  in  the  strains,  and  hence,  in  structures  of  the  kind,  the 
foundations  must  be  secure  from  settling,  and  the  pier  sup- 
ports accurately  on  level.  Under  these  circumstances,  .where 
long  spans  are  desirable,  the  continuous  girder  is  to  be  preferred 
to  a  succession  of  single  spans. 

If  the  greatest  negative  reaction  at  any  support,  as  E  (Fig.  87), 
is  greater  tnan  the  constant  positive  reaction  at  that  point  due 
to  the  dead  load,  the  girder  will  require  to  be  latched  or  held 
down  at  that  support. 

*  See  Appendix,  Art.  16,  where  this  comparison  is  made. 


198  CONTINUOUS    GIEDEE.  [CHAP.  XIII. 


CHAPTER  XIII. 


ANALYTICAL   FORMULAE   FOE    THE    SOLUTION    OF    CONTINOUS  GIEDEES. 

129.  Introduction. — As  we  have  seen  in  the  preceding 
chapter,  the  complete  and  accurate  determination  of  the  strains 
in  the  continuous  girder,  both  for  uniform  and  moving  loads,  is 
easy,  provided  we  can  find  the  moments  and  shearing  forces  at 
the  supports  for  the  various  states  of  loading,,  and  for  each  apex 
load.  Now  this  we  are  able  to  do  with  mathematical  accuracy, 
and  without  much  labor.  The  formulae  necessary  for  the  pur- 
pose, when  put  into  proper  shape  for  use,  are  neither  difficult  of 
application  nor  more  complicated  than  many  which  the  practi- 
cal engineer  is  often  called  upon  to  manipulate.  Since  the 
publication  of  Clapeyron's  paper  *  in  185 7,  in  which,  for  the 
first  time,  his  well-known  method  was  developed,  and  his  cele- 
brated "theorem  of  three  momenta"  made  known,  the  subject 
has  engaged  the  attention  of  many  mathematicians.  In  1862 
WinJder  f  first  developed  a  general  theory,  and  gave  general 
rules  fo?  the  determination  of  the  methods  of  loading  causing 
greatest  strains,  together  with  tables  for  the  maximum  values 
of  the  moments,  shearing  forces,  etc.,  for  various  numbers  of 
spans  of  varying  length.  In  the  same  year  Bresse  \  followed 
with  a  similar  work.  In  1867  WinHer  §  gave  a  general  ana- 
lytical theory,  and,  finally,  in  1873  Weyrauch  \  has  treated  the 
subject  with  a  degree  of  completeness  and  thoroughness  which 
leaves  but  little  to  be  desired.  He  discusses  the  subject  in  its 
most  general  form,  for  any  number  of  spans  of  varying  length, 

*  Clwpeyron — Calcul  d'une  poutre  elastique  reposant  librement  sur  des  ap- 
puis  inegalement  especes. — Compte  rendus,  1857. 

f  Beitrage  zur  Theorie  der  continuirlichen  Briickentrager —  Civil  Ingenieur, 
1862. 

\  Bresse — Cours  mechanique  appliquee.     Paris,  1862. 

§  Winkler— Die  Lehre  von  der  Elasticitat  und  Festigkeit.    Prag.  1867. 

H  Weyrauch — Allgemeine  Theorie  und  Berechnung  der  continuirlichen  und 
einfachen  Trager.  Leipzig,  1873. 


CHAP.  XT!!.]  ANALYTICAL  FORMULA.  199 

and  for  all  kinds  of  regularly  and  irregularly  distributed  and 
concentrated  loads — both  for  constant  and  varying  cross-section 
of  girder.  His  formulae  are  mathematically  exact,  and  for 
given  loading  are  free  from  integrals. 

The  above  is  but  a  very  imperfect  sketch,  and  we  have  named 
but  a  few  of  the  many  writers  who  have  been  occupied  with 
the  subject.  Clapeyrorfs  Theorem  above  alluded  to,  as  origi- 
nally given  by  him,  applied  only  to  uniform  load  over  whole 
length  of  girder,  or  over  an  entire  span.  But  as  early  as  JBresse's 
Treatise,  it  had  been  extended  to  include  concentrated  and 
local  loads  as  well,  and  Winkler  has  also  given  a  very  complete 
and  practical  discussion  of  the  subject. 

Notwithstanding  the  labors  of  these  and  many  other  mathe- 
maticians, there  seems  to  be  a  wide-spread  idea,  even  among 
those  who  are  supposed  to  have  considerable  familiarity  with 
mathematical  literature,  that  the  results  deduced  are  unpracti- 
cal. It  is  not  uncommon  to  meet  with  even  recent  publica- 
tions* in  which  it  is  stated  that  the  authorities  pass  over  such 
problems  with  "judicious  silence;"  that  the  mathematical  in- 
vestigations are  intricate,  and  the  formulae  deduced  trouble- 
some in  application  ;  that  even  a  a  partial  solution  of  the  prob- 
lem by  mathematical  calculation  is  attended  with  considerable 
difficulty,  and  that  a  complete  solution  for  the  bending  moment 
and  shearing  force  at  every  section,  under  moving  partial  and 
irregular  loads,  taxes  the  powers  of  the  best  mathematicians, 
and  is  well-nigh  impossible,  so  far  as  any  practical  application 
of  them  by  the  engineer  is  concerned."  How  far  such  ideas 
are  justified  may  be  seen  from  the  following  pages.  That  the 
authors  and  works  above  referred  to  can  only  be  read  by  good 
mathematicians  is  not  to  be  denied.  It  may  also  be  admitted 
that  the  subject  is  an  intricate  one,  and  when  treated  mathe- 
matically in  its  most  general  form  the  results  are  naturally  in  an 
unpractical  shape.  But  that  these  results  are,  therefore,  worth- 
less, or  that  the  formulae,  when  applied  to  any  particular  case, 
are  "  too  intricate  for  practical  use,"  by  no  means  follows. 

The  desirability  of  formulae  for  the  application  of  our  graph- 
ical method  as  developed  in  the  preceding  chapter ;  the  erro- 
neous ideas  prevalent  on  the  subject  which  we  have  just  noticed  ; 


*  Graphical  Method  for  the  Analysis  of  Bridge  Trusses :  Greene.     D.  Van 
Nostrand,  publisher,  New  York.  1875. 


200  CONTINUOUS    GIRDER  [CHAP.  XIII. 

and  the  deplorable  fact  that  the  "  authorities  "  do  but  too  often 
treat  the  subject  with  "judicious  silence,"  and  that,  therefore, 
there  exists  in  our  engineering  literature  no  collection  of  prac- 
tical and  useful  formulae  for  this  important  class  of  bridges, 
though  such  formulae  are,  and  have  been  for  years,  free  to  all 
for  the  asking, — all  these  facts  may  serve  as  apology  for  the  in- 
troduction of  the  present  chapter,  in  a  work  which  professedly 
treats  only  of  Graphical  methods.  The  apologies  of  those  who 
professedly  treat  the  subject  analytically,  and  have  yet  omitted 
such  formulae,  are  not  so  numerous. 

"We  propose  to  give  the  analytical  results  necessary  for  com- 
plete solution  of  a  girder  of  uniform  cross-section  over  any 
number  of  level  supports,  with  end  spans  any  desired  ratio  of 
intermediate ;  for  uniform  load  over  whole  length  from  end  to 
end  of  girder,  for  uniform  load  over  any  single  span,  and  for 
concentrated  load  in  any  single  span  at  any  point  of  that  span. 
These  three  cases,  as  we  have  seen  in  the  preceding  chapter, 
are  sufficient  for  the  complete  solution  of  framed  Bridge 
Trusses.* 

Many  of  these  results  are  here  given  for  the  first  time,  at  least 
in  their  present  shape,  in  any  published  treatise,  though,  as  re- 
marked, some  of  them  in  more  or  less  practical  form  have 
long  been  common  property  for  all  who  may  have  desired  to 
make  use  of  them. 

The  formulae  only  will  be  given,  in  such  shape  and  with 
such  illustrations  of  their  application  that,  we  trust,  they  will 
be  found  free  from  complexity,  and  of  considerable  practical 
importance.  In  the  Supplement  to  this  chapter  a  demonstra- 
tion of  the  formulae  is  presented. 

13O.  Notation. — The  notation  which  we  shall  adopt  is  as  fol- 
lows [see  Fig.  89,  PL  23]  : 

*  The  formulae  for  concentrated  loads  are  alone  all  that  is  really  necessary. 
Their  addition  gives,  as  we  have  seen  in  our  tabulation,  Art.  127,  the  strains 
for  uniform  load  also.  In  fact,  for  strict  accuracy,  only  single  isolated  loads 
should  be  considered,  as  the  results  given  by  the  formulae  for  uniform  load  are 
not  perfectly  accurate.  This  may  be  seen  from  the  well-known  fact  that,  for 
a  girder  fixed  at  one  end,  and  supported  at  the  other,  the  reaction  at  the  free 

5 
end  for  a  load  in  the  middle  is  -TO.  of  the  load,  while  if  the  same  load  were  uni- 

3  6 

formly  distributed,  the  reaction  is  ^ths,  or  ^  of  the  load.      The  difference, 

however,  for  any  practical  case,  where  there  are  a  number  of  panels,  is  very 
Blight. 


CHAP.  XIII.]  ANALYTICAL   FOKMULJE.  201 

Whole  number  of  spans  is  indicated  by  s  ; 

Hence,  whole  number  of  supports  is  s  +  1  —  numbered  from 
left  to  right. 

Number  of  any  support  in  general,  always  from  left  is  m. 

The  supports  adjacent  to  a  loaded  span  left  and  right  are 
indicated  by  r  and  y  +  1. 

When  extreme  end  spans  vary  in  length  from  the  interme- 
diate, they  are  always  denoted  by  n  I,  where  n  is  a  given  fraction 
or  ratio  for  any  particular  case.  Thus,  if  intermediate  spans 
are  all  70  feet  and  end  spans  50  feet,  I  =  70,  n  I  =  50,  and  n  — 


When  spans  next  to  ends  also  vary,  they  are  similarly  denoted 


All  other  spans  are  of  equal  length  and  denoted  by  I. 

The  length  of  the  span  in  which  the  load  is  supposed  to  be 
is  in  general  4>  where  the  value  of  r  for  any  particular  case  in- 
dicates the  number  of  the  loaded  span  from  left. 

A  concentrated  load  is  indicated  by  P. 

Its  distance  from  nearest  left-hand  support,  by  a. 

The  ratio  of  a  to  length  of  loaded  span  Zr,  is  It  =  —  . 

4 

Moment  at  any  support  in  general  is  MmJ  where  m  may  be 
1,  2,  3,  r,  r  4-  1,  s,  etc.,  indicating  in  every  case  the  moment  at 
corresponding  support  from  left. 

In  same  way  reaction  at  any  support  is  Rjn,  shear  Sm. 

At  supports  adjacent  to  loaded  span,  then,  we  have  Mr,  Mr+1, 
Rr)  R^,  Sr,  Sr+1,  for  the  moments,  reactions,  and  shears  at  those 
supports. 

A  dead  uniform  load  is  u  per  unit  of  length. 

A  uniform  live  load,  w  per  unit  of  length. 

w  4x,  then,  indicates  a  uniform  live  load  over  any  span. 

These  comprise  all  the  symbols  we  shall  have  occasion  to  use* 
By  reference  to  Fig.  89,  the  reader  can  familiarize  himself  with 
their  signification,  and  will  then  find  no  difficulty  in  under- 
standing and  using  the  following  formulae.  Certain  symbols 
which  we  shall  use  for  expressions  of  frequent  occurrence,  will 
be  best  explained  as  we  have  occasion  to  introduce  them. 

131.  <4  Theorem  of  Three  Moments.''  —  This  remarkable 
Theorem,  due  to  Clapeyron,  expresses  a  relation  between  the 
moments  at  any  three  consecutive  supports,  both  for  uniform 


202  CONTINUOUS   GIRDER.  [CHAP.  XIII. 

load  over  whole  length  of  girder  from  end  to  end,  and  for  uni- 
form load  over  the  whole  of  any  single  span.  It  may  be  writ- 
ten as  follows  : 


Mm  lm+  2  Mm+1  (lm+  lm+1)  +  Mm+2  Zm+1  =  [     +     +J. 

If  we  suppose  only  one  of  the  two  adjacent  spans  as  lm  to 
contain  the  full  live  load  -10,  while  all  the  spans  are  of  course 
covered  with  the  dead  load  u,  the  above  equation  becomes 


+1 


m    '          •*-"Am -t- 1  i_-iu    •     -1U  +  J.J     •     in +4  wia  +  i 

If  both  spans  bear  the  same  uniform  load  u  alone, 

]\Im  1+2  ]VTm  i\l    +  I     i"]  +  Hff        I       =  —  \f-\-f     "I 

If  the  spans  are  equal,  the  above  two  equations  become  re- 
spectively 

n..      l      O    » . 

MmZ  +  4Mm+1 1  + 
and 

Mm  I  +  4  Mm+1  I  +  Mm+2  I  —  r&?~* 

Now  in  every  continuous  beam,  whose  extreme  ends  are  not 
fixed,  two  moments  are  always  known,  viz.,  those  at  the  ex- 
treme supports,  which  are  always  zero.  Hence,  by  the  applica- 
tion of  this  theorem,  we  can"  form  in  any  given  case  as  many 
equations  as  there  are  unknown  moments,  and  then,  by  solving 
these  equations,  can  determine  the  moments  themselves. 

132.  Elx&mple—  Total  uniform  Load — all  Spans  equal. — 
Thus  let  it  be  required  to  find  the  moments  at  the  supports  for 
a  beam  of  seven  equal  spans,  uniformly  loaded  over  its  whole 
length.  The  moments  at  the  end  supports  Mt  and  My  are  zero. 
We  have  then,  by  the  application  of  the  last  equation  above,  the 
following  equations : 

Fdr  the  first  three  supports  1,  2  and  3,  m  —  1,  and 

4  Mg  I  4-  Ms  I  —  — ^— ,  or  4  3YT2  -f-  M3  =  — . 
For  supports  2,  3  and  4,  ny,  =  2,  and 

M2  +  4  M3  +  M4  =  " 


CHAP.   XIII.]  ANALYTICAL    FORMULAE.  203 

For  supports  3,  4  and  5,  m  —  3,  and 


or  since  in  this  case  the  moments  equally  distant  each  way 
from  the  middle  are  equal,  this  last  equation  becomes 

M3  +  4  M4  +  M4  •=  ?tf. 

a 

We  have  therefore  three  equations  between  three  unknown 
moments,  M2,  3YT3  and  M4,  and  by  elimination  and  substitution 
can  easily  find 


1  9 


If,  as  in  our  example  of  Art.  127  in  the  preceding  chapter, 
we  take  u  =  1  ton  per  ft.,  I  =  8'0  ft.,  then  u  1?  —  6400,  and  the 
moment  at  the  fourth  support  becomes  540.8.  If  the  height  of 
truss  is  ten  feet,  this  gives  [Fig.  88]  54.1  tons  strain  in  the 
upper  flange  A  a.  By  reference  to  our  tabulation,  Art.  12T, 
we  see  that  this  agrees  closely  with  strain  in  A  1  due  to  uni- 
form load,  found  in  a  manner  entirely  different,  viz.,  by  sum- 
mation of  the  strains  due  to  first  case  of  loading,  and  the  several 
loads  in  -the  span  itself,  and  serves  therefore  as  a  check  upon 
our  results. 

133.  Triangle  of  Momciit§.  —  For  the  benefit  of  the  practi- 
cal engineer,  who  may  object  to  the  algebraic  work  involved  in 
elimination  of  the  unknown  moments  from  the  equations  above, 
when  the  number  of  spans  is  great,  we  offer  the  following  tabu- 
lation, from  which  he  may  easily  and  directly  determine  the 
moments  at  the  supports  for  any  "desired  number  of  spans 
without  formulce  or  calculation. 

Thus,  if  we  were  in  the  above  manner  to  find  the  moments 
for  a  number  of  spans,  and  tabulate  our  results  as  given  in  the 
annexed  table,  an  inspection  of  the  table  will  show  us  that  we 
can  produce  it  to  any  extent  desired  without  further  calcula- 
tion. 


204 


CONTINUOUS    GIRDER. 


[CHAP.  XIIL 


MOMENTS   AT    SUPPORTS TOTAL  UNIFORM  LOAD ALL  SPANS  EQUAL. 

Coefficients  of  u  I2  given  in  triangle. 


The  Roman  numerals  along  the  sides  of  the  triangle  indicate 
tfye  number  of  spans,  and  the  horizontal  line  to  which  they  be- 
long give  the  moments.  Thus,  for  our  example  of  seven  spans 
just  worked  out,  we  have  the  extreme  moments  M!  and  M8  =  0, 

15 

M2  and  M7  =      —  u  Z2,  etc. 


Now,  a  simple  inspection  of  this  table  will  show  us  that  for 
any  even  number  of  spans,  as  VIII.,  for  example,  the  numbers  in 
the  horizontal  line  are  obtained  by  multiplying  the  fraction 
above  in  any  diagonal  column,  both  numerator  and  denomina- 
tor, by  2,  and  adding  the  numerator  and  denominator  of  the 
fraction  preceding  that. 


Thus, 


15  x2  +  11 


41 


142  x  2  +  104       388  '    142  x  2  +  104 


30  _  15  x 2 
388  ""  142x2 


12  x  2  +    9          33 
m   the   other    diagonal    column  ;     j^~2  +  104  ==  388      °r 

11  x  2  +  11   . 

=  :r77r~          T777  m  the  other  diagonal  column  ;  and  so  on. 
142  x  2  +  104 

For  any  odd  number  of  spans,  as  IX.,  we  have  simply  to  add 
numerator  to  numerator  and  denominator  to  denominator,  the 
two  preceding  fractions  in  the  same  diagonal  column. 


CHAP.  XIII.]  ANALYTICAL    FORMULAE.  205 

41   H-  15         56       33   +  12  30  +  15         45 

I  n  n  Q  —   —   __  Ol*    _          _    —    _ 

'  388  +  142  7  530'  388  +  142'  "  388  +  142  530' 
and  so  on. 

We  can,  therefore,  independently  of  the  theorem  and  analyt- 
ical method  by  which  the  above  results  were  deduced,  produce 
the  table  to  any  required  number  of  spans.* 

134.  Total  uniform  L,oad—  all  Span§  equal—  Reactions. 
—  The  moments  being  known,  the  reactions  at  the  supports  can 
be  very  easily  found. 

Thus,  the  reaction  at  the  first  or  last  support  is 


_ul      Ms 

':T~  T; 


at  any  other  support 


Thus,  in  our  example  in  Art.  127,  we  find 

56  161          _        137  143 


Hence,  the  shear  at  the  fourth  support  is 

56  161  137  143      .    .  71 


71 
or  when  u  I  =  80  tons,  -—     ul  =  4:0  tons. 


Multiplying  this  shear  by  1.414  (the  secant  of  the  angle  with 
vertical),  we  find  for  the  strain  in  diagonal  a  b  (Fig.  88)  due  to 
uniform  load  +  56.5  tons,  the  same  nearly  as  already  found  in 
our  tabulation. 

135.  Triangle  for  Reactions.  —  The  reactions  for  a  number 
of  spans  being  found,  and  tabulated,  as  above,  in  the  case  of  the 
moments,  we  sjiall  have  a  triangular  table  precisely  similar  to 
the  one  above,  in  which  the  same  rule  holds  good  for  odd  and 
even  numbers  of  spans. 


*  The  above  relations  between  the  moments  can  be  shown  analytically  to 
be  a  result  of  the  properties  of  the  well-known  "  Clapeyronian  numbers." 
For  the  table  above,  as  also  the  others  which  follow,  we  are  indebted  to  the 
kindness  of  Mr.  Mansfield  Merriman,  Instructor  in  Civil  Engineering  in  the 
Sheffield  Sci.  School  of  Yale  College.  They  are  given,  so  far  as  we  are  aware, 
in  no  treatise  upon  the  subject  yet  published. 


206 


CONTINUOUS    GIRDER. 


[CHAP.  xm. 


REACTIONS     AT     SUPPORTS TOTAL     UNIFORM       LOAD ALL      SPANS 

EQUAL. 

Coefficients  of  u  I  given  in  triangle. 


vni. 


VII. 


VII. 


VIII. 


"We  are  thus  able  to  find  both  moments  and  reactions  at  the 
supports  for  any  number  of  spans,  so  far  as  uniform  loading  is 
considered,  and  may  then  either  diagram  the  strains  in  the 
various  pieces  or  calculate  them  as  explained  in  Arts.  127  and 
128.  No  formulae  are  required.  Any  one  who  understands 
the  method  of  moments  as  applied  to  simple  girders  can,  by 
the  aid  of  the  two  tables  above,  find  accurately  the  strains  in 
every  piece  of  a  girder,  continuous  over  as  many  equal  spans  as 
is  desired,  and  uniformly  loaded  over  its  entire  length,  all  sup- 
ports being  on  the  same  straight  line. 

As  we  have  seen,  Art.  127,  this  is  one  of  the  cases  which 
must  be  considered  in  order  to  find  the  maximum  strains  in 
any  span,*  and  the  results  above  given  for  its  solution  will,  we 
trust,  be  found  by  the  practical  engineer  to  be  neither  "  com- 
plex "  nor  "  difficult  of  application." 

136.  Clapeyroiiian  Numbers. — In  the  analytical  discussion 
of  continuous  girders,  certain  numbers  having  many  remarka- 
ble properties  play  a  very  important  role. 

We  have  seen  that  the  theorem  of  three  moments  furnishes 
us  with  as  many  equations  between  the  moments  as  there  are 
moments  to  be  determined.  For  a  small  number  of  supports, 

*  See  note  to  Art.  129. 


CHAP.  XIII.]  ANALYTICAL   FORMULAE.  207 

these  equations  can,  be  solved  by  the  ordinary  rules  of  algebra ; 
but  for  a  great  number,  or  in  the  general  analytic  discussion  of 
any  number,  we  must  have  recourse  to  a  special  artifice.  Tims 
we  multiply  our  equations,  beginning  with  the  last,  by  numbers 
indicated  by  <?1?  <?2,  c3,  .  .  .  .  c^_i,  and  then  choose  these  num- 
bers such  that,  by  the  addition  of  all  the  equations,  all  the  JVC's, 
with  the  exception  of  Ml5  disappear.  We  thus  easily  determine 
M!  without  the  tedious  process  of  substituting  from  one  equa- 
tion to  the  other,  through  the  entire  list. 

The  following  relations  must  then  evidently  hold  between 
these  numbers,  as  is  evident  from  the  theorem  of  three  mo- 
ments of  Art.  131 :« 


2< 

\  (4-i  +  4)  +  <?2  i 
4-i  +  2  C2  (4_2  + 

Li  =  o. 

4-2  =  o. 

tfs-3  4  +  2  cs^  (4  +  4)  +  cs_,  4  =  o. 

If  the  first  number  is  chosen  at  will,  say  ±  13  the  other  num- 
bers can  be  found  from  these  equations. 

Now  in  the  present  case  of  all  spans  equal,  we  have  between 
any  three  of  these  numbers  the  relation : 

Cm-l  +  4  Cm  +  Gm  +  l  =  0. 

If  we  take  the  first,  ct  —  0,  and  the  next,  c%  =  1,  we  have 
for  the  others  the  following  values : 

Gl  =  0  c,  =  +  15  c,  =  -  780  CM  =  +  40545 

c2  =  -f  1        c5  =  -  56  <%  =  +  2911  cu  =  -  151316 

c3  zn  —  4        c6  =  +  200         c9  =  -  10864         c12  =  +  564719 

These  are  the  so-called  Clapeyronian  numbers.  They  alter 
nate,  as  we  see,  in  sign,  and  each  is  numerically  4  times  the 
preceding  minus  the  one  preceding  that.  We  shall  always  indi- 
cate these  numbers  by  the  letter  <?,  the  index  denoting  the  par- 
ticular one.  Thus,  c7  is  the  seventh  number,  counting  0  and 
1  as  the  two  first. 

No  table  of  these  numbers  is  needed.  The  index  being 
given,  any  one  can  write  down  the  series  for  himself,  till  he 
arrives  at  the  desired  number. 

137.  Uniform  Live  Load  over  any  single  Span — Moments 
at  Supports  of  Loaded  Span. — These  numbers  being  pre- 
mised, we  can  now  give  the  following  formulse  for  the  moments 
at  the  supports  r  and  r  +  1  of  the  uniformly  loaded  span  : 


208  CONTINUOUS    GIRDER.  [CHAP.  XII'. 

For  tl.e  left  support, 

M    ---  WP   |gr  gS-r+2  +  4.  6%_r  +  1     | 

L         c«+i        J 

For  the  right  support, 

Mr+1  =  -  -W  P   pr  gs-r  +  l 

These  formulae,  it  will  be  seen,  are  very  simple  an<}  easy  of 
application. 

Thus,  for  seven  spans,  load  over  the  fourth  from  left,  we 
have  s  =  7,  r  =  4,  and  hence 


1  72    F^   <"5   +   <°4   '4! 

--wP         ~ 


Both  moments  are  equal,  as  should  be  the  case  for  a  middle 
span.  Inserting  now  the  proper  values  for  the  Clapeyronian 
numbers  from  the  preceding  Art.,  we  have 

x—  56  4-152"!        615 


So  for  any  desired  number  of  spans,  the  values  of  r  and  s 
being  known,  the  corresponding  Clapeyronian  numbers  can  be 
easily  found,  and,  inserted  in  our  formulae,  give  us  at  once  the 
moments  at  the  supports. 

Turning  again  to  our  example,  Art.  127,  and  making  w  =  2 
tons,  and  I  =  80  ft.,  we  have  w  Z2  =  12800,  and  therefore  M4  = 
676,  and  dividing  by  depth  of  truss  —  10  ft.,  we  find  the  strain 
in  A  a  (Fig.  88)  67.6  tons,  nearly  what  we  have  found  by  the 
summation  of  the  strains  due  to  the  loads  Pt_4  in  our  tabulation. 

138.  Triangle  of  Moments—  Uniform  Live  Load  over  any 
single  Span.  —  If  from  the  above  formulae  we  find  the  moments 
at  supports  for  a  number  of  spans,  and  tabulate  as  before,  we 
shall  have  a  triangle  of  moments  similar  to  those  already  given, 
which  may  be  produced  to  include  any  desired  number  of 
spans.  We  have  only  to  observe  that  the  numerator  or  de- 
nominator of  any  fraction  in  the  table  follows  the  law  of  the 
Clapeyronian  numbers  —  that  is,  is  four  times  the  preceding  in 
the  same  diagonal  column  minus  the  one  preceding  that. 


CHAP.  XIII.]  ANALYTICAL   FORMULAE.  209 

MOMENTS    AT    SUPPORTS    OF     LOADED      SPAN. — UNIFORM    LIVE    LOAD 

OVER  ANY  SINGLE  SPAN. 
Coefficients  of  w  I'2  given  in  triangle. 


m,       -  153   x  4-  41         571  , 

Thus  for  seven  spans,          -      _  —  =  —,   and  so    on. 


The  triangle  above  gives  the  moments  for  uniform  load  over  any 
span,  both  right  and  left.  For  left  supports  we  have  simply 
to  count  the  span  from  right  to  left.  Thus  for  seven  spans  for 
load  in  the  sixth  span  from  left,  we  have  moment  at  left-hand 

627 
support  =  -  w  Z2,  counting  the  spans  from  left  to  right  in 


triangle.     For  the  moment  at  right  support  of  same  loaded 
span,  we  count  six  the  other  way  from  right  to  lejct,  and  find 

571 


13d.   Moments   at   Supports   of  Unloaded   Spans.  —  The 

triangle  and  formulae  above  give  the  moments  at  the  supports 
of  the  loaded  span  only,  both  positive  —  that  is,  always  tending 
to  cause  tension  in  upper  flange  and  compression  in  lower. 

If  m  represents  the  number  of  any  support  counting  from 
the  left,  the  moments  at  any  support  generally  may  be  found 
by  the  following  formulae  : 


When     m<r  +  l,    Mm  ==  \w  V  pm  Gs~r+*  +  ^m  Cg-r+11 

£g_l  +  4:  CB 

When      m>r     M^l^Z2       r  cs-m+2  +  <Vn  0M+i1 

<?s_i  +  4  <?a  J 


If  we  make  in  these  formulas  m  =  r  in  the  first,  and  m  = 
14 


210  CONTINUOUS   GIKDEK.  [CHAP.  XIII. 

r  4-  1  in  the  second,  we  obtain  the  formulse  of  Art.  137.  For 
any  other  support  left  of  r,  or  right  of  r  +  1,  we  have  only  to 
give  the  proper  values  to  m,  s  and  r  for  any  given  case,  and 
find  the  corresponding  Clapeyronian  numbers. 

14O.  Practical  Rule,  and  Table.  —  The  moments  at  the 
supports  of  the  loaded  span  having  been  found  by  the  formulse 
of  Art.  137,  or  the  triangle  of  moments  of  Art.  138,  instead  of 
using  the  above  formulse,  we  may  find  the  moments  at  the 
other  supports  as  follows  : 

For  all  supports  left  of  the  loaded  span:  Commencing  at 
the  left  end  support,  place  over  each  support  the  Clapeyronian 
numbers 

1  4  15  56  209  780,  etc. 
Take  the  last  number  thus  obtained,  before  reaching  the  left 
support  of  the  loaded  span,  as  a  common  denominator.  Then 
the  moment  at  the  left  end  is  of  course  zero.  At  the  second 
support  1,  at  the  third  4,  at  the  fourth  15,  at  the  fifth  56,  etc., 
all  divided  by  this  common  denominator,  will  express  the  frac- 
tional part  of  the  moment  at  the  left  support  of  the  loaded 
span,  which  the  moment  at  the  support  in  question  is.  For  the 
moments  at  the  supports  right  of  the  loaded  span,  proceed  simi- 
larly, only  count  from  the  right  end. 

Thus,  for  a  girder  of  ten  spans,  sixth  span  from  left  loaded  : 
The  moments  M6  and  M7  due  to  load  being  found,  suppose  we 
wish  the  moments  left  of  M6.  Commencing  at  left  end,  num- 
ber the  supports  1,  4,  15,  56,  209.  (Let  the  reader  draw  a  figure 
representing  the  case.)  The  number  209  is  the  last  before 
reaching  the  sixth  support.  We  take  this,  therefore,  for  a^com- 

mon   denominator.      Then  we  have  M^  =  0     M2  =  —  —  M6 

209 


So  for  supports  to  the  right  of  support  7,  we  have 


Remembering  that  M6  and  M7  are  both  positive,  and  that 
the  moments  alternate  either  way  from  these  supports,  we  find 
easily  the  proper  signs  for  the  moments  right  and  left. 

We  can  now,  therefore,  find  the  moments  at  D  and  E  due  to 
the  first  and  second  cases  of  loading  of  Fig.  87  (Art.  126). 


CHAP.  XIII.] 


ANALYTICAL  FORMULAE. 


Let  us  take  the  first  case.     For  load  on  A  B,  we  have  from 

/-rorv 

our  triangle  or  formulae  the  moment  at  B  =  w  P.   At  D, 


,          56         780 
then,  we  have  —  * 


56 


wl?. 


For  load  on  C  D,  we  have  at  once  from  triangle  the  moment 


Finally,  for  load  on  F  G,  we  have  for  moment  at  F,  from 


*  •    1      627 

triangle  =  ^r^-rr  w 


11644 


therefore  at  D,  ^  x  ^  «»  P  = 


45 


11644 


All  these  moments  at  D  are  positive  ;  we  have  therefore,  for 


the  first  case  of  loading,  the  total  moment  at  D  —  4- 


717 


w 


If  we  make  I  =  80  ft.  and  w  =  2  tons,  we  find  the  moment 
at  D  =  788,  and  dividing  by  10  we  obtain  78.8  tons  as  the 
strain  in  A  a,  Fig.  88,  corresponding  with  our  tabulation,  Art. 
127. 

Table  for  all  the  Moments. 

All  the  moments  may  be  found  from  a  simple  table  similar 
to  the  following,  which  will  be  found  perhaps  preferable  to  the 
triangle  of  Art.  138. 


TABLE  FOE  MOMENTS. — UNIFORM  LOAD  IN   SINGLE   SPAN. 

Support  counted  from  Left.  Denominator  A 


1 

2 

3 

4 

5 

6 

1 

0 

1 

4 

15 

56 

209 

I. 

0 

3 

12 

45 

168 

627 

II. 
III. 

0 

11 

44 

165 

616 

0 

41 

164 

615 

IV. 

0 

153 

612 

V. 

0 

571 

VI. 

I 

Spans. 

iA. 

1 

1 

2 

4 

3 

15 

4    - 

56 

5 

209 

6 

780 

7 

2911 

212  CONTINUOUS    GIEDEK.  [CHAP.  XIII. 

This  table,  it  will  be  observed,  can  be  produced  to  include 
any  number  of  supports  desired.  The  law  of  the  Clapeyroniaii 
numbers  runs  both  horizontally  and  vertically.  The  smaller 
table  gives  the  denominator,  the  larger  the  numerator  of  the 
coefficient  of  w  ft  for  any  case.  Thus,  for  seven  spans  we  have 
four  times  2911  =  11644  for  the  common  denominator.  For 
load  on  second  span  from  right,  moment  at  sixth  support  from 

627 
left,  we  have  then  directly  w  P  ;  for  fourth  support  from 

45 
left,  w  P,  the  same  as  above. 


For  load  in  fifth  span  from  right,  the  table  gives  us  at  once 

612 
and  +  ^—  -r-r  w  P,   for  supports   1,  2   and   3. 


For  the  other  supports,  since  if  now  we  were  to  continue  count- 
ing from  left  we  should  have  to  pass  a  loaded  support,  we 
must  count  the  loaded  span  from  left,  and  count  the  supports 
in  reverse  order.  For  fifth  support  from  right,  then,  the  num- 
ber required  is  at  intersection  of  III.  (instead  of  Y.)  and  5,  or 

(\~\  A 

5-j-j  w  P,  as  found  above.     Thus  the  tables  above  cover  all 
11644 

cases,  giving  supports  at  loaded  span  itself,  as  also  right  and 
left  of  this  span.  We  have  only  to  remember  to  count  sup- 
ports from  left,  and  loaded  span  from  right,  for  all  supports 
left  of  load,  and  inversely  for  all  supports  right  of  load.  [The 
reader  should  always,  when  using  Table,  make  a  sketch  of  the 
given  number  of  spans,  indicate  the  loaded  span,  and  number 
the  supportsl\ 

141.  Reaction*  at  Supports  —  Live  Load  over  single  Span. 
—  For  the  reactions  at  ends  of  loaded  span,  we  have 


For  reactions  at  extreme  ends,  when  end  spans  are  loaded,  or 


w 


When  any  other  spans  are  loaded,  or 

when  r  >  1  and  <  s,    R,  =  -  ???    Rs  +  1==_^?!. 


CHAP.  XIII.]  ANALYTICAL   FORMULA.  213 

For  all  other  reactions, 


Thus  for  load  covering  the  first  span  of  seven  spans,  we  find 
from  the  known  moment,  given  in  preceding  Art.,  for  the  fourth 
support, 

6  x  56  336 

R^n^wl  =  u^wL 

For  load  over  third  span  from  left, 

6  x  616  1      ,       6607 


For  load  on  sixth  span, 


Hence,  total  reaction  at  fourth  support  for  first  case  of  loading 

7213 
is  jpj-rr;  w  &•    In  the  same  way  we  can  find  the  reactions  at  the 


first,  second,  and  third  supports,  for  the  second  case  of  loading, 
as  shown  in  Fig.  87,  and  then  can  easily  find  the  shear  at  any 
support,  as  D,  by  taking  the  algebraic  sum  of  all  the  reactions 
and  loads  between  that  support  and  the  end. 

We  can  now,  therefore,  find  the  shear  and  moment  at  D,  and 
thus  determine  the  strains  in  the  span  D  E  for  both  cases  of 
loading,  as  given  in  our  tabulation,  Art.  127. 

142.  Triangle  for  Reactions  —  Single  Span  loaded.  —  If 
we  calculate  from  our  formulae  the  reactions  at  supports  of 
loaded  span,  for  a  number  of  spans,  we  can  tabulate  the  results, 
as  on  next  page,  in  a  triangle,  where  each  number  is  four 
times  the  preceding  minus  the  one  preceding  that,  all  in  the 
same  diagonal  column. 


214  CONTINUOUS   GIRDER.  [CHAP.  XIII. 

REACTIONS   AT  SUPPORTS — LIVE   LOAD   OVER   SINGLE   SPAN. 
Coefficients  of  w  I  given  in  triangle. 


This  triangle,  similar  to  the  preceding  one  for  moments, 
gives  the  moments  at  the  left  support  of  the  loaded  span,  when 
we  count  from  left  to  right.  Counting  the  other  way,  we  have 
the  reactions  at  the  right  support  of  the  loaded  span. 

Thus  for  six  spans,  fourth  span  from  left  loaded,  we  count 

1770 

four  from  left  in  horizontal  line  for  VI.,  and  find  w  I  for 

3120 

reaction  at  left  support.     For  reaction  at  right,  we  count  four 
also  from  right  end,  and  find  w  I. 


143.  Reaction§  in  unloaded  Spans—  Load  over  one  Span 
only—  Table.  —  The  formulae  of  Art.  141  for  the  reaction  at 
any  unloaded  support  are  sufficiently  simple  and  easy  to  apply  ; 
still  we  may,  if  thought  preferable,  also  draw  up  tables  for 
these,  to  be  used  in  connection  with  the  triangle  of  the  pre- 
ceding Art.  The  following  tables  give  the  coefficients  of  w  I 
for  the  reactions  not  adjacent  to  the  loaded  span.  The  denomi- 
nator of  the  fraction  is  to  be  taken  from  the  triangle  above  ;  the 
tables  referred  to  give  only  the  numerators. 


CHAP.  XIH.] 


ANALYTICAL  FORMULAE. 


215 


REACTIONS   AT   UNLOADED   SPANS. 
Supports  counted  from  left.  Supports  counted  from  right. 


1 

2 

3 

4 

5 

6 

1 

6 

24 

90 

336 

1254 

r. 

3 

11 

18 

72 
264 

270 
990 

1008 

ii'. 

66 



in1. 

41 
153 
571 

246 

918 

984 

IV. 

V. 



VI'. 

2131| 

VII'. 

B    -g 


6' 

5' 

4' 

a- 

2' 

1' 

I. 

II. 
III. 

1254 

336 
1008 

90 

270 

24 

72 

6 

18 

1 
~3~ 
11 

990 

264 

66 

IV. 
V. 
VI. 
VII. 





984 

246 
918 

41 
153 



571 
2131 

Tables  give  the  numerators  of  the  coefficients  of  w  L 
triangle  on  page  opposite. 


Denominators  from 


These  tables  may  be  carried  out  to  any  desired  extent  by  the 
law  of  the  Clapeyronian  numbers  in  the  vertical  columns. 

As  an  example  of  their  use,  take  seven  spans  load  in  fifth 
from  left,  that  is,  in  third  from  right.  (Make  sketch.)  From 
the  triangle  we  take  the  common  denominator  11644.  Then 
from  first  table  in  the  horizontal  column  of  III/  we  have  for 
left  end 


Ri- 


wl, Ra  =  — . 


66 


264_ 
11644' 


990 


11644 w  11644  ~  11644  ~  11644' 

For  supports  right  of  loaded  span,  we  must  take  the  second 
table,  and  look  in  horizontal  column  for  Y.  We  thus  obtain 

153  918 

^-11644^  *»-  ~  11644** 

We  can  now,  therefore,  either  by  our  tables  or  formulae,  or 
both,  find  the  moments,  reactions,  and  shearing  force  at  any 
support  for  both  cases  of  loading  given  in  Fig.  87.  The  reader 
will  do  well  to  take  the  example  of  Art.  127,  and  find  the  mo- 
ment and  shear  at  D  for  both  cases,  and  thus  check  our  results 
as  giren  in  Art.  127,  viz.,  +  788.2  ft.  tons  and  —  382.5  ft.  tons 
for  the  moments,  and  ±  14.63  tons  for  shear. 


216  CONTINUOUS    GLBDEE.  [CHAP.  XIII. 

144.  Concentrated  Load  in  any  Spun—  Moments  at  Sup- 
ports. —  It  only  remains  to  consider  a  concentrated  load  at  any 
point.  If  the  formulae  for  this  case  do  not  prove  to  be  too  com- 
plex or  intricate  for  practice,  we  may  consider  the  case,  so  far 
as  equal  spans  are  concerned,  as  fully  solved. 

We  have  seen  that  the  "  theorem  of  three  moments,"  so  far 
as  uniform  loads  are  concerned,  enables  us  to  solve  the  case 
thoroughly.  It  is  more  especially  as  regards  concentrated  or 
partial  loads  that  the  opinion  widely  prevails  as  to  the  impossi- 
bility of  obtaining  practically  useful  formulae  ;  and  this,  not- 
withstanding that  it  has  been  shown  by  Bresse,  WinJder,  Wey- 
rauch,  and  many  others,  that  the  theorem  of  three  moments 
can  be  extended  to  include  concentrated  loads  also. 

The  Theorem  as  thus  extended  is  as  follows: 

^-1  +  2  Mm    /^  +  Zm    +  Mm+1  lm  = 


Pm-l    a' 


where,  by  our  notation  (Fig.  89,  Art.  130),  a!  and  a  are  the  dis- 
tances of  Pm_i,  Pm,  from  the  nearest  left  supports.* 

By  the  aid  of  this  theorem,  we  are  able  to  deduce  the  follow- 
ing formulas  : 

For  moments  left  of  r,  and  including  support  r,  that  is 

when  m  <*•  +  !,    Mm  =  -  cm  A  Cs~r+2  +  A'  CS~T+I  . 

C&+1 

For  moments  right  of  r  +  1,  including  support  r  +  1,  or 

•u  TUT  A  Cp  +  A'  cI+l 

when  m  >  r,    Mm  =  —  ca_m+2  -  -  -  *&* 

CB  +  1 

In  these  formulae,  c  represents,  as  above,  the  Clapeyronian 
number,  and  A  A'  stand  for  the  following  expressions  : 

A  =  PZ(2£-3#J  +  P)    A'  =  PZ  (&-#), 
k  being  the  fraction  -,  or  the  ratio  of  the  distance  of  the  weight 

L 

P  from  the  left  support,  to  the  length  of  span. 

145.   Illustration  of  Application  of  above  Formulae.  — 

These  formulae  are  by  no  means  difficult  of  application.     Let 

*  For  demonstration  of  this  Theorem,  see  Supplement  to  this  chapter. 


CHAP.  XIII.]  ANALYTICAL   FORMULA.  217 

us  take  the  example  of  Art.  127  (Fig.  88),  where  P  =  40  tons, 
1=80  ft.,  and  a  becomes  10,  30,  50  and  70  ft.  respectively. 
First,  as  regards  the  expressions  A  A'  : 

These  become  in  the  present  case  3200  (2  &  —  3  ^  +  ^)  and 
3200  (k  —  %?)  respectively,  where  Jc  has  the  values  -J-,  f  ,  -|,  and  -J 
successively.  Now  as  the  denominator  is  in  each  term  always 
the  same,  in  the  first  8,  in  the  second  64,  in  the  third  512,  and 
only  the  numerators  of  the  values  of  Jc  vary  for  the  different 
positions  of  P,  we  may  put  these  values  of  A  and  A'  in  the 
forms 


or 

A  =  800  h  —  150  A2  4-  6.2305  A3, 
A'  =  400  h  -  6.2305  A3, 

where  A  has  successively  the  values  1,  3,  5  and  7,  for  P15  P2, 
P3  and  P4  respectively.  These  are  then  the  practical  formulae 
for  substitution  in  the  present  case. 

We  can  now  apply  the  formulae  for  M  above.  Thus,  sup- 
pose for  seven  spans  we  have  P3  in  the  fourth,  as  shown  by 
Fig.  88,  and  wish  the  moment  due  to  P3  at  the  fourth  support 
D.  Then  s  =  7,  r  =  4,  and  m  =  4,  and  we  have 

M4=-o4 


or,  referring  to  Art.  136  for  the  Clapeyronian  numbers, 
-56  A  +  15  A'       840  A  -  225  A' 


M4  =  -  15 


2911  2911 


Now  for  P3  we  have  k  =  -,  or  A  =  5,  and  therefore 

o 

A  =  1028.81     A'  =  1221.2. 
Hence 

=  202.4  ft.  tons. 

This  divided  by  10  =  height  of  truss  gives  tension  in  A  a  = 
20.2  tons,  nearly  what  we  have  already  found  in  our  tabulation, 
Art.  127. 

In  like  manner  we  may  easily  find  the  moment  at  D  due  to 


218  CONTINUOUS    GERDEE.  [CHAP.  XIII. 

every  weight,  or  by  giving  the  proper  value  to  m  in  our  for- 
mulae, we  may  find  the  moment  at  any  support  we  please. 

The  moments  at  the  supports  of  the  loaded  span  being  found, 
the  moments  at  the  other  supports  may  be  obtained  according 
to  the  rule  given  in  Art.  140  for  uniform  live  load  over  single 
span. 

146.  Triangle  of  Moments. — The  reader  may  also  by  the 
aid  of  the  formulae  above  form  a  triangle  similar  to  those  al- 
ready given,  containing  the  coefficients  of  P  I  for  the  moments 
at  the  supports  of  the  loaded  span. 

Thus  for  two  spans,  for  moment  at  left  support,  we  should 
obtain  0  and  J  [2  Jc  -  3  Z?  +  F]  P  I,  and  this  last  value  will 
run  down  the  right  diagonal  column  without  change,  except  in 
its  coefficient  J,  which  will  become  successively  T4^,  -J|-,  •££$, 
for  three,  four  and  five  spans  respectively.  For  three  spans 
we  shall  have,  0,  -^  [7  k  —  12  I2  +  5  %?]  P  Z,  and,  as  above, 
TS-  t2  ^  —  3  ^  +  ^]  P  I.  The  second  of  these  will  run  down 
the  second  diagonal  column  from  the  right  without  change, 
except  in  its  coefficient,  which  will  be  -fa,  ^V5??  etc.,  for  five 
and  six  spans. 

So,  for  four  spans  we  have 

0,  .fa  [26  Jc  -  45  &  -f  19  %?]  P  /,     ^  [7&  -  12  W  +  5  ^]  PI, 
^  [2  Jc  -  3  J<?  +  #]  P  I, 

for  moments  at  left,  for  load  in  1st,  2d,  3d  and  4th  span  re. 
spectively.  The  second  of  these  runs  down  the  third  diagonal 
column  from  the  right,  changing  coefficient  as  above. 

If  the  triangle  be  now  drawn,  and  these  expressions  properly 
inserted,  we  shall  observe  that  along  the  diagonal  columns 
sloping  down  and  to  the  left,  the  values  of  Jc  in  the  parenthesis, 
as  also  the  denominators  of  the  outside  fractions,  follow  the  law 
of  the  Clapeyronian  numbers.  The  numerators  of  these  out- 
side fractions  in  these  columns  remain  unchanged.  The  outer 
left  column  is  of  course  always  zero. 

Another  triangle  must  be  found  for  moments  to  the  right  of 
load,  and  then  the  moments  at  the  unloaded  supports  -may  be 
found  by  the  rule  of  Art.  140. 

All  the  moments  may  also  be  found  from  a  couple  of  tables 
formed  similarly  to  those  of  that  Art.  It  is  unnecessary  to 
give  such  tables  here.  From  the  above  the  reader  can  form 
them  for  himself,  if  desired.  The  formulae  for  moments  given 


CHAP.  Xin.]  ANALYTICAL   FORMULA.  219 

above  are  so  simple,  and  with  a  little  practice  so  readily  worked, 
that  tables  are  scarcely  needed. 

147.  Reactions  at  Support§  for  concentrated  Load  in 
§ingle  Span.  —  For  the  reactions  we  have  the  following  formulae  : 

1st.  Abutment  reactions. 

When  the  end  span  contains  the  load,  that  is,  when  r  =  1  or 


When  the  load  is  not  in  the  end  spans,  i.e.,  when  r  >  1  and 
r  <s, 

R  .         M2  Ms 

Rl-  -J-,     Rs+l  =  —. 

2d.  Reactions  at  supports  of  the  loaded  span  itself  (not  end 
span), 


3d.  For  all  other  reactions. 


• 


The  above  formulae,  in  view  of  what  has  been  said  in  Art. 
145,  are  sufficiently  simple  to  need  no  illustration. 

For  load  in  fourth  span  of  seven  spans  we  find  easily  for  the 
reaction  at  left  support, 

R4  =  g        [2911  -  3  Jc  -  6387  &  +  3479  #]. 


This  can  be  put  in  working  order  as  explained  in  Art.  145, 
and  the  reader  can  check  the  results  which  we  have  given  in 
the  example  of  Art.  127  for  himself. 

A  triangle  and  two  subsidiary  tables  for  the  reactions  at  the 
supports  of  loaded  spans  may  be  formed  similarly  to  the  tri. 
angle  and  tables  of  Arts.  142  and  143.  We  leave  this  for  the 
reader  to  accomplish  for  himself,  if  thought  desirable. 

148.  Shear  at  Supports  of  loaded  Span.  —  We  are  now  in 
possession  of  all  the  formulae  necessary  for  the  complete  solution 


220  CONTINUOUS   GIRDER.  [CHAP.  XIII. 

of  a  girder  over  any  number  of  supports,  all  spans  equal.  For 
any  desired  span,  we  can  find  the  maximum  positive  and  nega- 
tive moments  by  the  cases  of  Fig.  80,  as  also  the  moments  due 
to  various  positions  of  the  weight  P.  We  can  also  find  the  re- 
actions at  all  the  supports  due  to  these  cases.  From  the  reac- 
tions and  known  forces,  we  can  then  easily  find  the  algebraic 
sum,  or  shear  ,  at  any  support.  The  moment  and  shear  at  any 
support  due  to  any  case  of  loading  are,  as  we  have  seen,  the 
quantities  required  for  calculation. 

Now  it  is  not  necessary  to  find  all  the  reactions  in  order  to 
obtain  the  shear.  The  moments  at  the  supports  being  known? 
we  can  find  the  shear  directly. 

Thus,  for  concentrated  load  in  a  span  4  .(Fig.  89)  we  have  for 
any  point  x 

Mr  -  Sr  x  +  P  (4  -  a)  -  m  =  0, 

where  Sr  is  the  shear  at  the  left  of  the  loaded  span,  and  m  is 
the  moment  at  any  point.  We  see  at  once  that,  to  determine 
this  moment,  it  is  the  shear  that  we  wish,  and  not  the  reaction. 
For  a  uniform  load  we  have  similarly, 

Mr  -  Sr  x  +  ~  -  m  =  0. 
2i 

If  in  both  these  equations  we  make  x  =  1T)  m  becomes  M^, 
and  we  have 


where  q  =  P  (1  —  Jc)  for  a  concentrated  load,  and  q  =  -g£  for 

uniform  load. 

In  an  unloaded  span  at  the  left  support,  or  when  m  <  r, 
<]  disappears,  and  we  have 


For  the  shear  at  the  right  support  of  the  loaded  span  we  have 
simply  Sr  —  P  or  Sr  —  w  I,  and  hence 


CHAP.  XIII.]  ANALYTICAL   FORMULA.  221 

where  qr  —  P  k  for  concentrated  load,  and  q'  =  ^-^-  for  uni- 

2i 

form  load.     For  any  other  span  at  the  right  support 

S'      —  Mm  -  Mm-l 

^m-1 

Thus,  Sm  and  S'm  are  the  shears  at  any  support  just  to  right  and 
left  of  that  support  respectively.  The  reaction  at  any  support 
is  then  Rm  =  Sm  +  S'm. 

The  moments,  then,  at  two  successive  supports  being  known, 
we  can  readily  find  the  shear  at  any  support,  and  these  two, 
moment  and  shear,  we  repeat,  are  the  quantities  required  for 
calculation.  The  reactions,  and  the  tables  for  the  reactions 
above,  are  only  useful  as  enabling  us  to  find  the  shear.  It  is 
this  last,  together  with  the  moment  at  the  support,  which  gives 
us  the  moment  m  at  any  point  of  the  span  in  question,  as  is 
evident  from  the  above  equations.  It  is  only  in  the  case  of  the 
simple  girder  that  the  reactions  at  the  ends  are  the  same  as  the 
shears.  In  the  continuous  girder  only  the  latter  should  ~be 
used,  except  for  ends  of  end  spans,  where  the  two  are  identical. 
We  have  only  to  remember,  then,  that  the  shear  at  any  support 
is  the  algebraic  sum  of  all  the  reactions  and  loads  from  that 
support  to  the  nearest  extreme  end,  and  then,  knowing  these 
reactions  and  loads,  the  determination  of  the  shear  is  easy. 

We  might  give  tables  for  shears  directly,  as  above,  for  reac- 
tions; but  this  is  unnecessary.  Having  taken  the  reactions 
from  our  tables  already  given,  and  found  the  moments  either  by 
our  formulae  or  tables,  we  can  then  find  the  shears  both  by  means 
of  the  reactions  and  also  directly  from  the  moments  them- 
selves, and  thus  check  at  once  the  accuracy  of  our  determina- 
tion of  both.  From  what  has  already  been  given,  the  reader 
can  easily  construct  tables  of  shears  similar  to  those  alreadj^ 
given  for  reactions  for  himself,  if  desired. 

149.  Recapitulation  of  Formulae— Continuous  Girder 
over  any  Number  of  Level  Supports,  all  Spans  equal.*— 
For  notation,  see  Art.  130,  Mg.  89. 

*  These  formulae  are  given  in  similar  form  in  Winkler's  "Der  Lehre  von 
der  Elasticitaet  und  Festigkeit,"  Art.  144,  p.  122.  They  were  also  indepen- 
dently deduced  by  Mr.  Merriman,  to  whose  kindness  we  are  indebted  for  much 
of  this  chapter.  The  above  methods  of  tabulation  were  communicated  by 
him,  and  are  given  in  no  treatise  upon  the  subject. 


222  CONTINUOUS    GIRDER.  [CHAP.  XIII. 

1st.  Moments  at  supports. 
When  m<r  +  l,    M   =  -  C    A  gg~r+2  +  A/  GS-*+I 

when  m  >  r,    Mm  =  - 


in  which  &  =  *  and  A  =  P  J  (2&  -  3/P-f  /fc3),  A'  =  P  J  (^-  #) 

for  concentrated  loads,  and    A  =  A'  =  J  w  Z2   for  a  uniform 
load  over  any  one  span. 

2d.  Shear  at  the  supports. 

In  the  loaded  span,  to  the  right  of  the  left  support, 


To  the  left  of  the  right  support, 


S'r+1  =      »! 
«      r 


In  the  wMloaded  spans,  to  the  right  of  the  left  support, 


8     _ 


To  the  left  of  the  right  support, 

'        Mm  ~  Mm-1 


For  the  reaction  at  any  support,  Rm  =  S'm  +  Sm. 

3d.  Reactions. 

(a)  Abutment  reactions  : 

when  r  =  1,    R1=-?^  +  ^;  when  r  =  s,  Rs+1  =  -  —  s+/; 

when  y  >  1  and  <  s,   Rt  =:  —  ^     Rs+1  =  —  ???. 

^  ^ 

(5)  Eeactions  at  supports  adjacent  to  loaded  span  (when  this 
span  is  not  an  end  span)  : 


CHAP.  XIII.]  ANALYTICAL   FORMULAE.  223 

(<?)  All  other  reactions : 


Where  for  concentrated  load, 

A  =  P  I  (2  Jc-  3  &  +  IF)      A'  =  Pl(%-X*) 
q  =  P  (1  —  Jc)  qf  =  P  Jc. 

For  uniform  load  in  single  span, 

A  =  A'  =  J  w  P,      q  =  c[  =  \  w  I,    Jc  being  always  y,        and 

Cl  =  0  c,  =  15  c7  =  -  780        cw  =  40545 

c2  —  1  c5  =  -  56       c8  ==  2911  cu  =  -  151316 

c3  —  -  4       e6  =  209         c9  =  -  10864    c12  —  564719,  etc. 

We  give  also,  for  sake  of  completeness,  although  not  needed 
for  calculation,  the  formulae 

FOR  UNIFORM  LOAD  OVER  ENTIRE  LENGTH  OF  GIRDER.* 

Moment  at  any  support, 
Reactions  at  abutments, 


Reactions  at  other  supports, 


where,  as  before,     A  =  ^  wP,    q  —  \  w  I, 
and  the  numbers  indicated  by  5  are  as  follows  : 

&!  =  0                 &4  =  -  3  58  =  -  41  &a  =  -  571 

5a  =  1                £5  =  -  4  J9  =  -  56  518  =  -  780 

J3  =  1                56  =  +  11  J10  =  4-  153  etc.  , 

J7  =  +  15  ln  =  +  209 


*  The  above  equations  were  first  given  by  Mr.  Merriman,  in  the  Jour. 
Franklin  Institute,  April,  1875. 


224  CONTINUOUS    GIRDER.  [CHAP.  XIII. 

These  numbers  change  signs  by  pairs  alternately,  and  every 
other  one  follows  the  law  of  the  Clapeyronian  numbers.  In 
fact  those  with  the  odd  indices  are  those  numbers,  and  the  even 
ones,  commencing  with  0,  1,  3,  follow  the  same  law.  ' 

The  above  comprises  all  the  formulae  thus  far  given,  in  a 
shape  very  convenient  for  reference.  The  reader  who  has  fol- 
lowed attentively  our  explanation  of  their  use,  needs  nothing 
more  to  solve  the  case  of  equal  spans  completely.  The  expres- 
sions, however,  for  the  reactions  are  unnecessary.  As  we  have 
seen  from  Art.  148,  we  need  only  the  moments  and  shears  at 
any  support  in  practical  calculations.  The  practical  formulae 
necessary  and  sufficient  for  any  case  will  be  found  in  the  next 
Art. 

15O.  Girder  continuous  over  any  Number  of  Level  Sup- 
ports ;  Symmetrical  with  respect  to  the  Centre,  and  with 
two  variable  end  Spans  n  I  and  p  I  on  each  side.  —  [Fig. 
89.]* 

Moments  at  Supports  : 


Mm  = 


when  m>  r,   IV^  =  %*»  -  A  <*  +  . 

I      p  CB_!  +  2  (n  +  p)  ca 

Shear  at  Supports  —  loaded  span, 


4 
sWi = -  ~±V~  ~"~  "*"  ^' ' 

wwloaded  spans, 

•  B^=M,-M^t 

Mm  -  m^^ 


For  uniform  load, 

A  =  A'  =  J  w  fr ;     q  —  q'  =  \  w  lr 

*  Jour.  Franklin  Institute,  March  and  April,  1875. 


CHAP.  XIII.]  ANALYTICAL   FORMULA.  225 

For  concentrated  load, 

A  =  P  g  (2  k  -  3  &  +  #),     A7  =  P  £  (&  -  If), 

and          ^  =  P  (1  —  &),      £'  =  P  k,    k  being  always  •=-. 

^r 

The  quantities  denoted  by  c  are  also  as  follows  : 


_ 


+  12  j9)  -  n  (14  + 


j>  (52  +  45  j>)  +  ^(52  +  60  p) 

T 

_  -p  (194  +  168  p)  -  n  (194  +  224  p) 
-J~ 

p  (724  ^  617^)  +  w  (724  +  836^) 
c8  —  -  —  —  —  —  ,  etc., 

following  the  law  of  the  Clapeyronian  numbers. 

151.  Application  of  the  above  Formulae.  —  The  for- 
mulse  of  the  preceding  Art.  comprise  in  a  most  compact  form  all 
the  formulae  hitherto  given,  and  are  all  that  is  necessary  for  the 
complete  solution  of  any  practical  case. 

Thus,  by  making  p  —  unity  and  retaining  only  71,  we  have 
the  case  of  a  girder  with  variable  end  spans  n  I,  of  different 
length  from  the  others,  which  latter  are  all  equal  and  repre- 
sented by  I.  The  reader  will  find  no  difficulty  in  using  the 
above.  For  any  particular  case,  when  w  or  P  and  Z,  k,  n  and 
p  are  given,  A,  A7,  q  and  q'  can  be  easily  found,  and  the  prob- 
lem is  solved.  If  n  and^)  be  both  unity,  we  have  the  formulae 
for  all  spans  equal.  The  expressions  for  Mm  will  then  reduce 
to  those  already  given  in  Art.  144.  Thus,  in  Art.  145  we  have 
already  found  for  seven  equal  spans,  I  =  80  ft.,  load  P  =  40 
distant  50  ft.  from  left  ;  the  moment  M4  =  202.4.  Now,  from 
our  formulae  above,  we  find  for  MS  making  m  —  5,  5  =  7, 
r  =  4;  M5  =  272.8. 

Then  by  our  formulae  for  shear,  S4  =  +  14.12,  or  nearly 
15 


226  CONTINUOUS    GIKDER.  [CHAP.  XIII. 

what  we  have  assumed  in  Art.  145.  We  may  also  find  the 
same  shear  by  finding  the  algebraic  sum  of  the  reactions  at 
ABC  and  D  from  the  formulae  of  Art.  147.  This  is  more 
tedious,  and,  as  we  see,  unnecessary.  The  moments  can  be 
easily  found,  and  then  the  shear  obtained  directly  from  these. 

We  must  bear  in  mind  that  lr  always  denotes  the  span  the 
load  is  upon,  whether  nl,pl,  or  I,  while  lm  is  any  span  in  gen- 
eral, according  to  the  value  of  m. 

152.  Continuous  Girder  with  fixed  ends.—  It  is  worthy  of 
remark  that  if  n  be  made  zero  in  the  formulas  of  Art.  149,  we 
have  a  girder  with  fastened  ends  and  variable  end  spans  p  I. 
If  in  addition^?  is  unity,  then  all  the  spans  become  equal.    We 
must,  however,  remember  that  when  we  thus  make  n  =  0,  the 
number  of  spans  is  s  —  2  instead  of  s,  as  before/  and  the  end 
spans  are  p  I  ;  the  end  supports  are  also  2  and  s  instead  of  1 
and  s  +  1. 

153.  Examples.  —  As  illustrations  of  the  use  of  the  formulae 
of  Art.  150,  we  give  a  few  examples. 

Ex.  1.  A  beam  of  one  span  is  fixed  horizontally  at  the  ends. 
What  are  the  end  moments  and  reactions  for  a  concentrated 
weight  distant  k  \from  the  left  end? 

Here  the  two  outer  spans  of  three  spans  are  supposed  zero. 
Therefore,  5  —  2  =  1,  and  s  =  3.  The  left  end  is  2  instead  of 
1,  and  the  right  end  3.  Hence,  r  —  2,^>  =  0,  and  n  =  1  in  the 
formulae  of  Art.  150.  We  have,  then, 

d  =  0,     c2  =  1,     CQ  =  —  2,    £4  =  4>          and  hence, 


or,  inserting  the  values  of  c  above, 

M2  =  1(2A-A'),    M3=-1(A-2A'). 

For  a  concentrated  load,    A  =  P  Z2  (2  k  -  3  #  +  #), 
and      A'  =  P  P  (&  -  ;&»).     Hence,  M2  =  P  I  (k  -  2Z?  +  .#), 
and  M3  =  P  I  (1$  -  Vs). 

For  the  reaction  at  the  left  end,  which  is  in  this  case  the 
same  as  the  shear,  we  have 


CHAP.  Xm.]          ANALYTICAL  FORMULAE.  227 

52  =  M2-Ms  +  P  (1  _  k)   or   S2  =  P  (1  -  3  #  +  2  #), 

f 

53  =  M3-Ms  +  P  k         or         S3  -  P  (3  #  -  2  #). 

^ 

For  a  load  anywhere,  we  have  simply  to  give  the  proper  value 
to  k,  and  we  have  at  once  the  reactions  and  moments.  Thus, 
for  a  load  at  \  the  span  from  the  left,  Tc,  =  £,  and 

s2  =  fjp,  SS  =  AP;  M2  =  &Pi,mi  =  &Pi 

For  a  load  in  centre,  k  —  J,  and 


[Compare  Supplement  to  Chap.  VII.,  Arts.  16  and  IT.] 
Ex.  2.  J^r  #  uniform  load  over  the  same  beam,  what  are  the 

end  moments  and  reactions  f 

We  have  simply  to  introduce  the  proper  values  of  A  and  A' 

for  this  case,  and  we  have  at  once 

Mg  —  T^-  w  Z2  =  M3    and    S2  =  S3  =  £  w  I 

Ex.  3.  A  girder  of  three  equal  span's  is  "  walled  in  "  at  the 
ends,  and  has  a  concentrated  load  in  the  first  span.  What  are 
the  moments,  shears,  and  reactions  at  the  ends  and  intermediate 


In  this  case,   s  —  2  =  3,  and  hence    s  =  5,    r  =  2,   n  —  0, 
j9  =  1,  and  therefore 

C2      A  i5  +  A'  c4 
=  T        *4  +  2*     ' 

frA^  +  A'fr  ^AC^  +  A^ 

2       c4  4-  2  c?5  ^      c4  +  2  c5 

also,         ct  =  ,  0  c2  =  1,  <k  —  —  2,  c4  =  7,  c5  =  —  26,  etc. 

Inserting  these  values  and  the  values  of  A  and  A'  for  con- 
centrated load,  we  have 

P  7  91 


=  -       (3  &  -  3  #),  M5  =       (3  ^  -  3 

4bO  4:0 


228  CONTINUOUS    GIEDEE.  [CHAP.  XIII. 

Observe  that  the  moments  are  positive  at  each  end  of  the 
loaded  span  and  alternate  in  sign  from  that  span,  varying  as 
the  numbers  1,  2,  7,  or  as  the  numbers  c  [Art.  140].  A  positive 
moment  always  denotes  compression  in  lower  fibre.  For  the 
shears  we  have,  then,  from  Art.  150, 

S2  =  -?-  (45  -  99  #  +  54  #%   S'8  =  -^  (99  &  -  54  Jf). 
4o  45 

Ss  =  ^  (27  ^  -  27  #),  S'4  =  ^  (-  27  /fc2  +  27  #), 

4:0  4:5 

S4  =  ^  (-  9  #  +  9  #),  S'5  =  ^  (9  #>  -  9  #). 

4-0  4:0 

For  the  reactions,  then, 

Ra  =  S2,  R3  =  S'8  +  S8  =  ^  (126  #  -81.&3), 

4-O 

R4  =  S'4  +  S4  =  ?-  (-  36  &  +  36  #),   R5  =  S'5. 

4:0 

Observe  that  the  reactions  as  also  the  shears  are  positive  at 
the  supports  of  the  loaded  span,  and  alternate  in  sign  from 
those  supports.  A  positive  shear  or  reaction  acts  always  up- 
wards. Disregarding,  then,  for  the  present,  the  weight  of  the 
beam  itself,  it  would  have  to  be  held  down  at  first  pier  from 
right  end. 

If  the  weight  is  in  the  centre  of  first  span  from  left,  Jc  —  J,  and 


36  _      9 

360    '       -  360    '  360    '  ^  ~  860 


..«  _p  B  R  _ 

^  ~          '  *•  -  '     4  '  ^  ~         * 


The  reactions  add  up  to  P,  as  they  should. 
If  P  =  100  tons,  and  1  =  16  ft.,  we  have 

M2  =  237.5  ft.  tons,  M3  =  87.5,  M4  =  -  50,  M5  =  25  ft.  tons. 
R2  =  60  tons  RS  =  47.5,     R4  =  -  10,    R5  =  2.5  tons  ; 

S2  =  60  tons,  S'8  =  40,       S8  =  7.5,        S'4  =  -  7.5, 

S4  =  -2.5,  S'5  =  2.5. 

Ex.  4.  A  beam  of  five  spans,  free  cut  ends  ;  centre  and  adja- 
cent spans  100  ft.,  end  spans  each  75  ft.,  has  a  uniform  load 
extending  over  the  whole  of  the  second  span  from  left.  What 
are  the  moments  at  the  ends  and  supports  f 


CHAP.  Xin.]  ANALYTICAL   FORMULAE.  229 

Here  s  =  5,  n  =  f  ,  p  —  1,  r  =  -2  ;  therefore,  from  Art.  150, 
.2_A.5   +  A'*4  ^A.2+   A'.3 

*  ^4     +    1  4,  Z  C4     +    J  C5 

and         <?!  —  0,  £2  =  1,  <?3  =  —  f,  c4  —  13,  £5  =  —  48.5. 

w  Z3 
Since,  then,  A  =  A'  =  —  —  for  uniform  load,  we  have 


!  =  0,  M2  =  -g-VV  ™  P>   M3  -  Tf  I*  w  ?,    M4  =  -  y  35^  w 

M  ^  0. 


If  the  load  is  two  tons  per  ft.,  w  1?  =  20,000,  and 
Mt  =  0,  M2  =  414.7,  M3  =  1036.6,  M4  =  -  279.5,   M5  =  79.7. 

Find  the  shears  and  reactions  at  each  support. 

Ex.  5.  A  beam  of  four  equal  spans,  has  the  second  span  from 
left  covered  with  full  load.  What  is  the  moment  and  shear  at 
left  of  load  f 

Ans.         M2  =  Tg^j-  w  I*,         S2  =  JJJ-  w  I. 

What  at  right  of  load  ? 

Ans.         M3  =  3^-  w  I,         S'3  =  %  Jf  w  I. 

What  are  the  formulae  for  concentrated  load  ? 

5o 

TV/r    — r'r  Z>        1 0  Z»2    i    K   L3~|  ~p  7 

J-TXo   f  /*»     I    *     tv  ~~"   -L^    'v      ~i      *-^    *v     I    ^  v* 

OD 

S2  =  ^  [56  -  58  ^  +  3  t?  -  ^3], 

S'8=  [58^-3^  +  ^]  ^>- 

Examples  might  be  multiplied  indefinitely. 

The  above  is  sufficient  to  show  the  comprehensiveness  of  our 
formulae,  and  the  ease  with  which  results  may  be  obtained, 
which,  by  the  usual  methods,  would  require  long  and  intricate 
mathematical  discussions.  The  points  of  inflection  and  the 
deflection  may  also  in  any  case  be  easily  determined,  and  gen- 
eral equations  similar  to  the  above  deduced,  but,  as  we  have 
seen,  the  above  are  sufficient  for  full  and  complete  calculation. 

154.  Table*  for  Moment*. — From  the  formulae  of  Art.  150 
we  can  easily  find  the  moments  for  both  uniform  and  concen- 
trated load  in  a  single  span  for  various  numbers  of  spans.  If 
these  results  are  tabulated  we,  shall  obtain  tables  from  which 


230 


CONTINUOUS   GIEDEE. 


[CHAP,,  xm. 


the  moments  may  be  at  once  taken.  The  formulae  for  the 
shears  are  so  easy  when  for  any  case  the  moments  are  known, 
that  it  is  unnecessary  to  give  tables  for  these. 

The  reader  will  do  well  to  make  himself  perfectly  familiar 
with  the  formulae  by  calculating  the  moments  for  various  cases, 
and  comparing  with  the  following  tables.  We  give  the  prac- 
tical case  of  variable  end  spans  n  I  and  equal  intermediate 
spans  I. 

TABLE  FOB  MOMENTS — UNIFORM  LOAD   OVER  ANY   SINGLE   SPAN. 
Coefficients  'of  w  I2  from  talk.    End  spcuns  n  1. 

Supports  counted  from  left. 


1        2 

3 

4 

5 

6 

0         n* 

(2  +  3n)n3 

(7      8«)»» 

(26  +  3071)^3 

(97  +  11271)7*3 

I. 

0      1+271 

(l  +  2ra)  (2  +  27i) 

(1+270(7+871) 

(l+27i)(26  +  80n) 

etc. 

II. 

0      5  +  0  n 

(5  +  6ra)(2  +  3/i) 

(5  +  6  n)  (7  +  8  n) 

etc. 

III. 

0    19  +  2271 

(19  +  22»)(2  +  2») 

(19  +  227i)(7  +  87i) 

IV. 

0   71  +  8271 

(71+83n)(2  +  8n) 

etc. 

V. 

0  365+  304  n 

etc. 

VI. 

Number  of 

spans. 

One-fourth  of  Denominator. 

1 

2 

3 

3  +  8  Ti  +  4  7ia 

4 

12  +  2871  +  16  n» 

5 

45  +  104  71  +  60  7i2 

6 

168  +  3837*  +  224  r 

I* 

The  above  table  can  be  easily  extended  to  include  any  num- 
ber of  spans.  It  is  precisely  the  same  as  the  table  of  Art.  140, 
and,  in  fact,  includes  that  table.  We  have  only  to  make  n  =  1 
and  we  have  at  once  the  table  for  equal  spans.  Suppose  we 
take  five  spans,  load  in  second  from  right.  From  the  smaller 
table  we  have  at  once  for  the  denominator  of  the  coefficient 
of  w  Z2,  4  (45  +  104  n  +  60  n2).  Then  from  the  other  table 
we  have  at  support  1  from  left  Mx  =  0, 

(l+2n)wP 


at  support  2, 


4  (45  +  104  ?i 


ANALYTICAL   FORMULA. 


CHAP,  xm.] 
,  at  support  3,       M3  •—  (1  4-  2  n)  (2  +  2  n)  w 


231 


,  etc. 


4  (45  4- 104  n  +  60 

If  the  load  were  in  second  span  from  left,  and  supports  to 
right  of  load  were  required,  we  have  simply  to  count  the  sup- 
ports the  other  way  in  the  table. 

Thus,  3YE6  =  0,  3YC3  =  — - — - —       — - .,  or  same  as  3VT2 

in  first  case,  etc. 

TABLE   FOR   MOMENTS CONCENTRATED  LOAD  IN  ANY  SPAN,  k  =   — . 

End  spans  n  1.     Coefficients  of  P  \Tfrom  table. 


a  =  a' 

(P 

y 

0=0' 

<i 

V' 

11. 

3  +  271 

1 

2n 

1 

2  +  271 

3  +  4» 

II'. 

III. 

9  +  10/1 

2  +  2  n 

3  +  471 

2  +  271 

7  +  871 

12  +  1471 

III'. 

IV. 

33  +  38% 

7  +  871 

12  +  14w 

7  +  87i 

26  +  307* 

45  +  5271 

IV. 

v. 

123  +  142  n 

26  +  3071 

45  +  52n 

26+3071 

97  +  11271 

168  +  19471 

V. 

VI. 

459  +  53071 

97  +  11271 

168  +  194  n 

97+11271 

362  +  41871 

627  +  72471 

VI'. 

VII. 

etc. 

etc. 

etc. 

etc. 

etc. 

etc. 

VII'. 

Number  of 
spans. 

Denominator  A. 

2 

3 

3  +  87i  +  4n2 

4 

12  +  2S?t  +  167i2 

6 

45  +  10471+60712 

6 

168  +  388  »+  224712 

7 

etc. 

The  above  tables  give  only  the  moments  at  the  supports  of 
the  loaded  span.  The  Roman  numerals  L,  II.,  III.,  etc.,  denote 
the  number  of  this  span  from  left,  and  I'.,  II'.,  III'.,  the  num- 
ber of  the  loaded  span  from  right.  The  expression  for  the 
moment  at  left  support  is 


M 


=  P  lr^  [7'  &  -  3  ff 


CONTINUOUS    GIRDER.  [CHAP.  XIII. 

For  the  right  support, 


where  the  expressions  for  0,  0',  A,  7,  7'?  £  £',  a,  a',  are  to  be 
taken  from  the  tables  and  inserted. 

Thus,  for  five  spans  load  in  second  span  from  right,  or  third 
from  left,  we  have  at  once  A  =  46  +  104  n  +  60  n\  For  the 
moment  at  the  left  support,  to  find  0,  we  must  take  horizontal 
line  for  III.,  and  thus  find  (2  +  2  *).  For  7',  /3'  and  a'  we 
must  take  II'.,  and  find,  therefore,  3  +  4  n,  2  +  2  n  and  3  +  2  n. 
Hence,  moment  at  left  support  is 


For  moment  at  right  support,  we  must  take  line  II'.  for  6'  and 
line  III.  for  7,  ft  and  a,  and  hence 


Since  the  span  in  question  is  not  an  end  span,  1^  =  1  and 


For  a  load  in  an  end  span,  use  the  formulae  of  Art.  150. 
For  a  load  in  middle  span  of  five  spans,  i.e.,  third  span  from 
each  end,  we  have 


+-    n  n)  *~3  (7+8  n)  *'  + 

M  '=«•= 


When  Jc  =  -=,  both  these  moments  become,  as  they  should,  equal 

2i 

for  any  assumed  value  of  n,  as  the  reader  may  readily  prove  by 
insertion. 

For  the  other  moments  not  adjacent  to  the  loaded  span,  the 
rule  of  Art.  140  holds  good. 

Thus,   Mt  =  0,  M,  =  I  M,,  M8  =  J  M,,  M4  =  -J  M,,  etc., 

P  "  " 


CHAP.  Xin.]  ANALYTICAL  FOKMUL^EJ.  233 

and  similarly  on  the  other  side, 

M8  =  0,    M^  =    *    Mr+1,  Ms_2  =  Jz?  M;^,  etc. 
C/r+1  UI+l 

We  must  remember  always  to  give  the  proper  signs  to  the 
moments,  viz.,  positive  for  extremities  of  loaded  span,  and 
alternating  each  way  from  these  for  the  others.  From  the 
formulae  of  Art.  140  we  can  then  easily  find  the  shear  at  any 
support. 

155.  Continuous  Girder—  I^evel  Supports  —  Spans  all  dif- 
ferent— Oeneral  Formulae.*  —  The  preceding  formulae  com- 
prise the  case  of  one  or,  at  most,  two  variable  end  spans.  We 
give  below  the  general  formulas  for  all  spans  different.  These 
formulas  include  all  the  others  as  special  cases.  Thus,  if  we 
make  all  spans  equal,  we  have  the  formulae  of  Art.  149.  If  end 
spans  \  and  ls  are  made  zero,  and  we  take  the  number  of  spans 
equal  to  s—  2,  and  first  support  2,  we  have  the  continuous  gir- 
der with  fixed  ends,  in  which  the  intermediate  spans  may  or 
may  not  be  equal,  as  we  choose.  If  we  make  ^  —  0  or  Z8  =  0 
alone,  and  s  —  1  =  No.  of  spans,  we  have  a  continuous  girder 
fixed  at  one  end  only.  In  short,  the  formulae  comprise  the 
entire  case  of  level  supports.  They  are  as  follows  : 

Let  s  =  number  of  spans,  4=length  of  loaded  span,  7c  =  —  , 

Ly 

a  being  distance  of  load  from  left  support  ;  l^  4>  4  •  •  •  •  4-i>  fc 
the  lengths  of  the  various  spans  counting  from  left. 
Then,  when  m  <  r  +  1, 

TVT     —  /«        A 

m4<-i 
when  m  >  r, 

M      ±A  A  er  +  B  cr+l 

A™-m  —  "'s—  m+27  —  —  ;  —  o    (1      i  —  7  -  \  —  * 

V_i  tfs-i  +  2  (4  +  ^.j  ca 


For  the  shear  at  supports  of  loaded  span, 

gr  =  Mr-Mr+1  +  g    SWi^M      -Mr  + 

6r  lt 

For  -zmloaded  spans, 


*  These  formulas  were  first  given  by  Mr.  Merriman,  and  may  be  found  In 
the  London  Phil.  Magazine,  Sept.,  1875. 


234  CONTINUOUS    GIEDEE.  [CHAP.  XIII. 

For  the  reaction  at  any  support,  Rm  =  S'm  +  Sm.     In  which 

we  have  always  Jc  =  -  ,  q  =  P  (1  —  #),  q'  =  P  Jc. 
i>r 

A  =  P  lr2  (2  k  -  3  &  -f  If)  and  B  =  P  I*  (k  -  F)  for  concen- 
trated load  ;  and  q  =  q'  =  ^  wliy  and  A  =  B  =  %  wl*  for  uni- 
form load  entirely  covering  any  one  span. 

Also  for  G  and  d  we  have  the  following  values  : 


=  1, 

2  ft  + 
" 


_  4  ft  4-  4)  (4  +  4)  -  V 

—         ""' 


or,  generally, 


_  4:  ft  +  d)  (4-1   +   4-2)  ~ 


74-7  7 

"s— m+3     i     ^s— m+2         J         ^s— m+3 


d5  =  —  2  d± 

or,  generally, 

dm=  —2d 


As  an  illustration  of  the  use  of  the  above  formulae,  let  us 
take  three  unequal  spans,  load  in  the  first.     Then  s  =  3,r  =  ~L, 

Tc,  =  y.   For  moment  at  second  support,  m  —  2,  or  m  >  ^ ;  hence 
^i 

M  --  /7         A  ^i  +  B  g2 

^  * 


CHAP.  Xm.]  ANALYTICAL  FOKMUL^E.  235 

But    Ci  =  0,   C2  =  1, 


hence,  since  B  =  P  If  (k  —  If), 

_  2  P  y  (*  -  Jf)  (4  +  4) 
*  (4  +  4)  (k  +  4)  -V 

If  in  this  we  make  \  =  4?  we  have  the  extreme  spans  equal, 
and  then 

M  _  a  P  y  (*  -  #)  ft  +  4) 

*  (4  +  4)2  -  V 

If  we  make  in  this,  again,  \  =  4,  we  have  for  all  spans  equal 


just  what  we  should  have  from  Art.  149. 
For  the  reaction  at  the  end  support,  we  have 


or,  since  Mt  =  0, 


For  all  spans  equal,  or  1^=1^=1^  =  1^  this  reduces  to 

S!  =  Z  (15  -  19  Jc  +  4  #»), 
15 

as  we  should  have  found  from  Art.  149. 

Ex.  1. — A  beam  of  one  span  is  fixed  horizontally  at  the 
right  end  ;  what  are  the  reactions  and  the  moments  for  concen- 
trated load  ? 

Here  5  —  1=1  or  s  =  2,  r  =  1,  4  =  0,  and  from  the  for- 
mulae of  Art.  155,  cl  =  0,  c2  =  1,  and  <h  =  0,  d%  =  1,  d$  =  —  2, 

,  A  GI  +  B  c2        B  c% 


.=  -2  +  P  (!-&)  =      (2  - 
'a  =  ?  (3  *-£>). 


236  CONTINUOUS   GIRDER.  [CHAP.  XIII. 

Ex.  2.  —  A  'beam  of  three  spans  of  25,  50  and  40  feet  respec- 
tively is  fixed  horizontally  at  the  right  end,  and  has  a  concen- 
trated load  of  10  tons  at  12  feet  from  the  third  support  from 
left.  What  are  the  moments  at  the  supports  f 

Here  I,  =  25,  ^  =  50,  4  =  40,  4  =  0,  P  =  10,  k  4  =  12, 
Jc  =  0.3,  5-1  =  3.  s  =  4,  4  =  0  and  r  =  3.  Also,  cl  =  0, 
c2  =  1,  GS  =  —  3,  c4  =  12.25  and  ^  =  0,  ^  =  1,  4  =  —  2, 
<Z4  =  6.4,  d5  =  -  32.4. 

When,  then,  m  <  4, 


Inserting  &  =  0.3  and  the  values  of  <?, 

for  m  =  1,  M!  =  0  ;  m  =  2,  M2  =-  8.20  ;  m  =  3,  M3  =  24.62  ; 
for  n  =  4, 


M4  =         (_  3  A  +  12.25  B)  =          (6.25£  +  9#-  15.25  #), 

or  M4  =  42.29  ft.  tons. 

Find  the  shears.  Also  moments  and  shears  for  uniform  load 
over  third  span. 

Ex.  3.  —  A  beam  of  four  spans  \  =  80,  13  =  100,  13  =  50, 
14  =  40  ft.,  free  at  the  ends,  has  a  load  of  10  tons  in  the  sec- 
ond span  at  40  ft.  from  left.  What  are  the  moments? 

Here  s  =  4,  ^  =  0,  c*  =  1,  <?3  =  —  3.6,  c4  =  19.6,  c5  =  -83.7, 
dj.  =  0,  4  =  1,  <&  =  -  3.6,  d,  =  10.3,  4  =  -  41.85,  r  =  2. 

For  m  <  3,      Mm  =  -  ^-  (A  ^4  +  B  4), 


m>3, 
Hence,  Mj  =  0, 


Us  <17  *  -  30-9  *  +  13-9  *>  = 


M5  =  0. 

Find  the  shears.     Also  find  the  moments  and  shear  for  uni 
form  load  over  second  span. 


CHAP.  XIII.]  ANALYTICAL   FORMULA.  237 

156.  Thus  we  see  that,  as  in  Art.  150,  a  few  short  and  sim- 
ple formulae,  which  may  be  written  on  a  piece  of  paper  the  size 
of  one's  hand,  are  all  that  we  need  for  the  complete  solution  of 
any  case  of  level  supports — whether  the  spans  be  all  equal  or 
the  end  ones  only  different,  or  all  different ;  whether  the  girder 
merely  rest  on  the  end  supports  or  be  fastened  horizontally  at 
one  or  both  ends.  We  have  only  to  remember  that  a  positive 
moment  causes  tension  in  upper  flange  at  support,  and  there- 
fore compression  in  lower;  inversely  for  negative  moment. 
Also,  that  a  positive  shear  acts  upwards,  and  a  negative  shear 
downwards.  Also,  that  both  moment  and  shear  are  positive  at 
supports  of  loaded  span,  and  alternate  in  sign  both  ways.  This 
is  all  that  we  need  to  form  properly  the  equation  of  moments 
at  any  apex,  and  determine  the  quality  of  the  strains  in  flanges 
and  diagonals.  We  can  thus  solve  any  practical  case  of  framed 
continuous  girder  which  can  ever  occur  with  little  more  diffi- 
culty than  in  the  case  of  a  simple  girder. 

Thus,  for  the  span  DE  (Fig.  87)  we  have  only  to  find  the 
moments  at  D  and  E  due  to  every  position  of  P  in  the  span 
D  E,  and  the  corresponding  shears  at  D.  These  once  known, 
and,  as  we  have  seen,  they  can  be  easily  obtained  from  our 
formulae,  we  can  find  and  tabulate  the  strains  in  every  piece 
due  to  each  weight,  as  shown  in  Art.  127.  An  addition  of  these 
strains  gives,  then,  the  maxima  of  each  kind  due  to  interior 
loading. 

We  have,  then,  to  find,  in  like  manner,  the  strains  due  to  the 
two  cases  of  exterior  loading  as  represented  in  Fig.  87.  From 
the  four  columns  thus  obtained,  we  can  deduce  the  dead  load 
strains,  and  then  finally  the  total  maximum  strains  of  each  kind 
for  every  piece.  [See,  for  illustration  of  the  above,  Art.  127.] 

Thus,  the  whole  subject  is  solved  with  the  aid  of  but  four 
simple  formulas,  and  for  a  problem  generally  considered  impos- 
sible by  reason  of  its  "  complexity,"  our  results  will,  we  trust, 
be  found  sufficiently  simple  and  practical. 

In  view  of  the  fact  that  the  necessary  formulae  for  practical 
computations  have  been  often  given  in  the  later  works  of 
French  and  German  authors,  although  perhaps  never  before 
in  so  compact  and  available  a  shape  as  above,  it  is  indeed  sur- 
prising that  they  should  have  been  so  completely  ignored  by 
English  and  American  writers. 

The  tables  and  formulae  which  we  have  given  will,  we  trust, 


238  CONTINUOUS    GIRDER.  [CHAP.  XIII. 

bring  the  subject  fairly  within  the  reach  of  the  practical  engi- 
neer, and  should  they  be  the  means  of  calling  more  general  at- 
tention to  this  important  class  of  structures,  will  not,  we  hope, 
be  considered  as  out  of  place  in  the  present  treatise. 

For  the  influence  of  difference  of  level  of  the  supports,  as  well 
as  for  variable  cross-section  and  the  relative  economy  of  the 
continuous  girder,  see  Arts.  17  and  18  of  the  Appendix. 


ART.  1.]  SUPPLEMENT   TO    CHAP.  XIH.  239 


SUPPLEMENT  TO  CHAPTER  XIII. 

DEMONSTRATION   OF   ANALYTICAL  FORMULAE   GIVEN   IN   TEXT. 

IN  the  following  we  shall  give  the  complete  development  of  the  general 
formulae  of  Art.  155.  As  these  formulae  include,  as  we  have  seen,  all  the 
others  as  special  cases,  it  is  sufficient  to  show  how  they  are  obtained  in 
order  to  enable  the  reader  to  deduce  all  the  others. 

1.  Conditions  of  Equilibrium. — In  the  rth  span  of  a  continuous 
girder,  whose  length  is  IT  (see  Fig.),  take  a  point  o  vertically  above  the  rih 


support  as  the  origin  of  co-ordinates,  and  the  horizontal  line  o  I  as  the  axis 
of  abscissas.  At  a  distance  x  from  the  left  support  pass  a  vertical  section, 
and  between  the  support  and  this  section  let  there  be  a  single  load  Pr 
whose  distance  from  the  support  is  a. 

Now  all  the  exterior  forces  which  act  on  the  girder  to  the  left  of  the 
support  r  we  consider  as  replaced,  without  disturbing  the  equilibrium,  by 
a  resultant  moment  Mr  and  a  resultant  vertical  shearing  force  Sr.  This 
moment  is  equal  and  opposite  to  the  moment  of  the  internal  forces  at  the 
section  through  the  support  r ;  while  the  vertical  force  is  equal  and  oppo- 
site to  the  shear. 

Not  only  over  the  support,  but  also  at  every  section,  the  interior  forces 
must  hold  the  exterior  ones  in  equilibrium,  and  therefore  we  have  the  con- 
ditions : 

1st.  The  sum  (algebraic)  of  all  the  horizontal  forces  must  be  zero. 

2d.  The  sum  (algebraic)  of  all  the  vertical  forces  must  be  zero. 

3d.  The  sum  (algebraic)  of  the  moments  of  all  the  forces  must  be  zero. 

Thus,  for  the  section  #,  we  have  from  the  third  condition 

2  M  =  Mr  -  Sr  x  +  Pr  (x  -  a)  -  m  =  0    .     .     .     .     (1) 
where  m  is  the  moment  at  the  section.    From  this  we  have 

m  =  Mr  —  Sr  x  +  Pr  (x  —  a) .     (2) 

If  in  this  we  make  x  =  1&  m  becomes  Mr+i,  and  we  thus  have  for  the  shear 
just  to  the  right  of  the  left  support  of  the  loaded  span 


24:0  SUPPLEMENT    TO    CHAP.  XIII.  [ART.  2. 

For  an  unloaded  span  the  weight  P  disappears,  and 


For  the  shear  just  to  the  left  of  the  right  support  of  loaded  span, 


For  unloaded  span,  the  weight  P  disappears,  and 


S'm  is  then  the  shear  to  the  left  of  any  support  m,  and  Sm  that  to  the 
right.  The  reaction  at  any  support  is  therefore 

Rm  =  S'm  +  Sm. 

These  are  the  formulae  already  given  in  Art.  148. 

2.  Equation  of  tlie  Elastic  Line.—  We  can  now  easily  make 
out  the  equation  of  the  elastic  line  for  the  continuous  girder  of  constant 
cross-section,  or  constant  moment  of  inertia. 

The  differential  equation  of  the  elastic  line  is,* 


where  E  is  the  coefficient  of  elasticity,  and  I  the  moment  of  inertia. 
If  now  we  insert  in  (3)  the  value  of  m,  as  given  in  (2),  we  have 

d?  y  _  Mr  —  Sr  x  +  PT  (x  —  a) 
dH?~  El 

Integrating  f  this  between  the  limits  x  =  0  and  .?,  and  upon  the  condition 

that  x  cannot  be  less  than  a,  the  constant  of  integration  ~~  =  tr  =  the  tan- 
gent of  the  angle,  which  the  tangent  to  the  deflected  curve  makes  with  the 
horizontal  at  r  ;  and  we  have,  since  we  must  take  the  /  Pr  (x  —  a)  simul- 
taneously between  the  limits  x  —  a  and  z  for  x  —  0  and  x  ; 

dy_         2Mrg-Sta?  +  Pr(g-a)« 

cTx-tr+  2EI  '     (3'^ 

If  we  take  the  origin  at  a  distance  hT  (see  Fig.)  above  the  support  r, 
then  integrating  again,  the  constant  is  hT,  and  we  have 


which  is  the  general  equation  of  the  elastic  curve.     If  in  this  we  make 

*  See  Supplement  to  Chapter  VII.,  Art.  12. 

f  Notice  that  when  x  =  0,  a  —  0,  and  hence  (x  —  a)  —  0  also. 


ART.  3.] 


SUPPLEMENT   TO    CHAP.  XIII. 


x  —  lr,  y  becomes  hr+i.     If  also  we  put  -  =  &,  or  a  =  k  Zr,  and  insert  also 


for  Sr  its  value  as  given  in  (2  a),  we  find  for 


_  _         2  Mr  Zr  -f  Mr+1  Zr  -  Pr  Zr2  (2  *  -  3 


..(5) 


We  see,  then,  that  the  equation  of  the  curve  is  completely  determined, 
when  we  know  Mr  and  Mr+1,  the  moments  at  the  supports.  These,  as  we 
shall  see  in  the  next  Art.,  are  readily  found  by  the  remarkable  "theorem 
of  three  moments,"  already  alluded  to  in  Art.  144. 

3.  Theorem  of  Tliree  moments. 


In  the  Fig.  we  have  represented  a  portion  of  a  continuous  girder,  the 
spans  being  li  Z2  .  .  .  /r?  etc.,  and  Ihe  supports  1,  2  ...  r,  etc.  Upon  the 
spans  IT-I  and  lr  are  the  loads  Pr_i  and  Pr,  whose  distances  from  the  near- 
est left-hand  supports  are  k  /r_i  and  k  lr  ;  k  being  any  fraction  expressing 
the  ratio  of  the  distance  to  the  length  of  span. 

The  equation  of  the  elastic  line  between  Pr  and  the  r  +  1th  support  is 
given  by  (4),  and  the  tangent  of  the  angle  which  the  curve  makes  with  the 
axis  of  abscissas  is  given  by  (3  a).  If  in  (3  «)  we  substitute  for  ST  its 
value  from  (2  a),  and  for  tr  its  value  from  (5),  and  make  at  the  same  time 


x  —  lr,  then  -£  becomes 
a  x 


1,  the  tangent  at  r  +  1,  and  we  have 


2  M 


r+1 


-  pr 


Remove  now  the  origin  from  o  to  ft,  and  we  may  derive  an  expression 
for  tr  by  simply  diminishing  each  of  the  indices  above  by  unity ;  therefore 

7,  T,  ^  I-  -I 


-- 


^  Zr_i  +  2  Mr  ^-Pr,!  Zr2_!  (*- 


r—  1  - 

Now,  comparing  these  two  equations,  we  may  eliminate  the  tangents, 
and  thus  obtain 

Mr_i  ?r_!  +  2Mr  (7r_!  +  Zr)  +  Mr+l  t  = 


_  6  E 


(2  yfc- 


which  is  the  most  general  form  of  the  theorem  of  three  moments  for  a 
girder  of  constant  cross-section. 

When  the  ends  of  the  girder  are  merely  supported,  the  end  moments  are, 
of  course,  zero.  Then,  for  each  of  the  piers,  we  may  write  an  equation  of 
the  above  form,  and  thus  have  as  many  equations  as  there  are  unknown 
moments. 

16 


242 


SUPPLEMENT    TO    CHAP.  XIII. 


[ART.  4. 


4.     Determination    of  the   Moments— Support§    all  on 

level.— When  all  the  supports  are  in  the  same  horizontal,  the  ordinates 
hij  hi,  #r,  etc.,  are  equal;  and  hence  the  term  involving  E  I  disappears, 
and  we  have  simply 

Mr_!  Zr_!  +  2  M?  (Zr_i  +lr)  +  Mr_!  Zr  = 

Pr_i  IJ-i  (Tc  -  F)  +  Pr  V  (2  k  -  3  #»  +  &3), 
as  already  given  in  Art.  144. 

Now  let  s  =  number  of  spans,  and  let  a  single  load  P  be  placed  on  the 
rth  span.  [PL  23,  Fig.  89.] 

From  the  above  theorem,  since  M!  and  Ms+1  are  zero,  we  may  write  the 
following  equations : 

2M2  (I,  +  Z2)  +  M3Z2  =  0; 

M2  Z2  +  2  M3  (Z3  +  13)  +  M4  Z3  =  0. 


Zr) 


Mr+1  ^  = 

=  A 


Mr_!  Zr-l  +  2  Mr  (Zr_ 
PZr2(2&-  3 

Mr  Z,.  +  2  Mr+1  (Z,.  +  Zr+i)  +  Mr+2  Zr+i  = 

P  Z,-2  (&  -  P)  =  B. 


Ms_2  Zs_ 


Ms  ^  =  0; 


=,  0. 


(6) 


2  Ms_!  (7S_2 
Ms_i  Zs_!  +  2  Ms  (Zs_!  + 

The  solution  of  these  equations  can  be  best  effected  by  the  method  of 
indeterminate  coefficients,  as  referred  to  in  Art.  136. 

Thus  we  multiply  the  first  equation  by  a  number  c2,  whose  value  we 
shall  hereafter  determine,  so  as  to  satisfy  desired  conditions.  The  second 
we  multiply  by  c3j  the  third  by  c4,  the  rth  by  Cj.+i,  etc.,  the  index  of  c  cor- 
responding always  to  that  of  M  in  the  middle  term.  Having  performed 
these  multiplications,  add  the  equations,  and  arrange  according  to  the  co- 
efficients of  M2,  M3,  etc.  We  thus  have  the  equation 

[2  C2  (li  +  Z2)  +  C3  Za]  M2  +  [ca  Z2  +  2  Ca  (la  +  Z3)  +  C4  Z3]  M3  +    .    .   . 
+  [Cr_i  Zr_i  +  2  Cr  (Zr_i  +  ZT)  +  CT+I  Zr]  Mr  +   .   .   .   . 

+  [cs_2  Zg_2  +  2  cs_i  (Zs_2  +  Zs_i)  +  CB  Zs_i]  Ms_! 

+  [Cg_i  Zs_i  +  2  C8  (Zs_!  -f  ?„)]  Ms  =  A  Cj.  +  B  Cr+1. 

Now  suppose  we  wish  to  determine  Ms.  We  have  only  to  require  that 
such  relations  shall  exist  among  the  multipliers  c  that  all  the  terms  in  the 
first  member  of  the  above  equation,  except  the  last,  shall  disappear.  We 
have  then  evidently,  for  the  conditions  which  these  multipliers  must 
satisfy, 

21  c2  (Zi  +  Z2)  +  Z3  Z2  =  0  ; 

c2  Z2  +  2  c3  (Z2  +Z3)  +  c4  Z8  =  0  ; 


Cr_l  Zr_!  +  2  C,.  (?r_!  +  Zr)  +  CT+I  Z,.  =  0  J 
Cs_2  Z8_2  +  2  C8_!  (Zs_2  +  Z8_!)  +  CB  Z?_i  =  Oj 


ART.  4.]  SUPPLEMENT   TO    CHAP.  XIH.  24:3 

while  for  Ms  we  have  at  once, 

__Acr 


-  _ 

Cs_i  Zs_i  +   2  Cs  (la_-L  +   Zg  )  CB+i  I* 

If,  in  like  manner,  we  should  multiply  the  last  of  equations  (6)  by  the 
number  d2,  the  last  but  one  by  d3,  the  rth  by  <4_r+1,  etc.  ;  then  add,  and 
make  all  terms,  except  that  containing  M2,  equal  to  zero  ;  we  should  have 
the  conditions  : 

2ds(k  +  Zs_i)  +  <ZsZB-i=0; 

d*  4-1  +  2  d3  (4_i  +  4-2)  +  dt  4_2  =  0  ; 


2  <4-r+2  (4  +  4-l)  +  ^s-r+3  4-1  =  0  ; 


4_i  Z3  +  2  ds_i  (lt  +  Za)  +  <4  Z2  =  0  ; 
while  for  the  moment  we  have 


The  values  of  M2  and  Ms  are  thus  given  in  terms  of  the  quantities 
A  and  B  and  c  and  d. 

A  and  B  depend  simply  upon  the  load  and  its  position  in  the  rth  span. 
Thus  A=PZr2(2£-3&2  +  P),  B  =  PZr2(fc-F). 

As  for  the  multipliers  c  and  d,  they  depend  only  upon  the  lengths  of  the 
spans,  and  need  only  satisfy  the  conditions  above.  Hence,  assuming 
d  =  0,  ca  =  1,  and  dl  =  0,  d*  —  1,  we  can  deduce  the  proper  values  for 
all  the  others.  Thus, 

d  =  0,  d,  =  0, 

ci  --1,  d*  =  1, 


0        Za  +  ?3  Za  ^  0    T    Zg_i  -f  Zs_2         ,   Zs_ 

C4  =  —  2  C3  —  -  --  C2  -,  »4  =  —  2  rfs  -  5  -   —  rfa  ^  — 
Zs                  Z3  Z8-2  4- 

0       Zs  +  Z4  Zs  ,  n  J   4-2  +  4-3        *     4- 

C6  =  —  2  C4  —  =  -  —  C3  —  ,  a*,  =  —  4  Ci4  -  -  -  —  «3    j 


O  ^    ^4  +  ^5         ^    ^4 

Ce  =  —  2  Cs  —  =  -  —  C4  —  , 

65  ZB 


4- 


etc.,  etc.  etc.,  etc. 

Now  from  equations  (6)  we  see  at  once  that  M3  =  c3  M2,  M4  =  c4  M2, 
etc.,  or,  universally,  when  n  <  r  +  1, 

.      .     (7) 


Also  taking  the  same  equations  in  reverse  order,  M^  =  d*  M8,  Ms_2  = 
d*  MB,  etc.,  or,  universally,  when  n  >  r, 

...     (8) 


SUPPLEMENT   TO    CHAP.  XIII.  [ARTS.  5,  6. 

Equations  (7)  and  (8)  are  the  general  equations  given  in  Art.  155,  which, 
as  we  have  seen,  include  the  whole  case  of  level  supports. 

5.  Uniform  Load. — For  uniform  load  the  same  equations  hold 
good.     We  have  only  to  give  a  different  value  to  A  and  B. 

Thus,  for  several  concentrated  loads  we  should  have 

A  =  S  P  Zr2  (2  lc  -  3  F  +  &3). 

For  a  uniform  load  over  the  whole  span  I  r,  let  w  be  the  load  per  unit 
of  length,  then 

ri  ri 

2P  =    I  wd  a;  or  since  a  =  lc  Zr,  2  P  =    /    wlTdlc. 

Jo  Jo 

Inserting  this  in  place  of  2  P  above,  and  integrating,  we  have 

A  =  B  =  i  w  Zr3. 

Thus  the  equations  of  Art.  155  hold  good  for  concentrated  and  uniform 
load  in  any  span,  for  any  number  and  any  lengths  of  spans. 

The  above  formulae  were  first  published  in  an  article  on  the  Flexure  of 
Continuous  Girders,  by  Mansfield  Merriman,  C.E.,  in  the  London  Phil, 
Magazine,  Sept.,  1875. 

6.  Formulae  for  the  Tipper. — The  expressions  for  the  reactions 
in  this  case,  already  given  in  Art.  120,  may  be  easily  deduced.     The  solu- 
tion is  tedious  by  reason  of  lengthy  reductions,  but  the  process  of  deduc- 
tion is  simple.  - 

The  construction  in  this- case  is  indicated  in  Fig.  83,  PL  22.  We  sup- 
pose, as  shown  there,  a  weight  upon  the  first  span  only.  Under  the  action 
of  this  weight  the  beam  deflects,  and  one  centre  support  falls  and  the 
other  rises  an  equal  amount.  Thus,  if  we  take  the  level  line  as  reference, 
A2  =  —  ha.  Moreover,  the  reactions  at  these  two  supports  must  always  be 
equal. 

We  have,  then,  as  representing  this  state' of  things,  A2  =  —  A3,  and  calling 
the  supports  1,  2,  3  and  4,  we  have  from  Art.  1,  since  Mj  —  M4  =  0,  and 

Zi  =  Za, 

R,  =  Si  =  S,  =  -  ^  +  P  (1  -  jfe), 


•B    —  a'  ms 

**-St  =  --!7' 

These  reactions  will  evidently  be  known,  if  we  can  determine  the  mo- 
ments. 

Let  Yr  =  6  E  I  F^r  -  ftr-l  +  ftr-ftr  +  ri         Then    the    gen_ 

L       4r-l  TT         J 

eral  equation  of  three  moments  of  Art.  3  becomes,  when  we  neglect  Pr,  that 
is,  suppose  only  the  first  span  loaded  : 

Mr_i  Zr_i  4-  2  Mr  (7r_!  +  Zr)  +  Mr  +  1  Zr  =  -  Tr  4  Pr-l  V-l    (^  ~  *")• 

This    expresses  a  relation  between   the  moments  at  three  consecutive 


AET.  7.]  SUPPLEMENT   TO    CHAP.  XIII.  245 

supports  for  load  between  the  first  two.     Let  r  —  1  =  1,  or  r  =  2.     Then, 
since  Mj  =  M4  =  0,  we  have 

2Ma(fa  +  fa)+M3fa  =  -Ya  +  Pfaa(;&-A;3)=R     .     .     (10) 
where  R  stands  for  convenience  equal  to  the  expression  on  right. 

Let  r  —  1  =  2,  or  r  =  3.     Then  the  weight  disappears,  and  since  fa  =  fa, 
M3fa  +  2M3(fa  +  fa)  =  -Y3     ......      (11) 

From  (11)  we  have 


But  since  R2  must  always  equal  R3,  we  have  from  (9) 

M.-M.  »_».--  8M.  =  _Pit 

fri  ta 

Substituting  (12)  in  (10),  we  have 


_  R  k  „  a  Y,  (fa  +  fa) 

3fa'+8fafa  +  4fa' 
Substituting  (12)  in  (13), 

_fafa«Pft-Ya(fa  +  2fa) 

8fa'+8fafa  +  4fa' 
From  (14)  and  (15),  we  have  then 

-  Y3  -  R  =  I,  fa  P  k. 
Insert  in  this  the  value  of  R  from  (10),  and 

Ya  -  Y3  =  P  I?  (k  -  If)  +  P  I,  fa  T&  =  P  (IS  k  -  fa2  ¥  +  fa  fa  A). 
Now  in  the  present  case  hi  =  0,     7it  =  0,     and  h2  —  —  A3,  and  since  also 
fa  =  fa, 


and  Y2  =  6Eir3--+-l-     That  is,  Y2  =  -  Y3. 

Hence,  from  our  equation  above, 


Substituting  these  values  of  y2  and  y3  in  (11)  and  (13),  we  can  obtain  at 
once  M2  and  M3,  which  finally  substituted  in  eq.  (9),  will  give  us  the  re- 
actions as  already  given  in  Art.  120,  when  we  put  n  I  in  place  of  fa. 

7.  In  similar  manner  we  can  solve  other  problems.  Thus  —  what  are  the 
reactions  for  a  girder  continuous  over  three  supports,  the  two  right-hand  ones 
resting  upon  an  inflexible  body  which  is  pivoted  at  the  centre  ? 

This  is  the  case  of  the  tipper  when  raised  at  the  centre  so  that  the  ends 
just  touch,  and  then  subjected  to  a  load  at  any  point  of  first  span  —  the 
other  end  not  being  latched  down,  so  that  it  rises  freely,  as  though  without 
weight  of  its  own. 


246  SUPPLEMENT   TO   CHAP.  XTTT.  [ABT.  7. 

In  this  case  we  have  from  (9),  since  now  M3  =  0, 


By  the  conditions  Ra  must  equal  R3,  hence 


2  I,  M2  +  Z2  M3  =  -  ll  I,  P  TQ    .     .    .     *•    .     .     (16) 
From  the  equation  of  three  moments  above,  we  have,  making  r  —  1  =  2, 
or  r  =  3,  since  then  P  disappears, 

M2Z2  =  -  Y3     .     .  ,•>    ,     ....     .     .     (17) 

ivr          Ts 

or  M'=-T 

Substituting  this  value  of  M2  in  (16),  we  find 

Y,  =  ^*;  hence  M.  =  -  ^ 

2  ii  +  ta  <«  61  +  ta 

and  therefore,  at  once, 


Putting  w  Z  in  place  of  Z2,  we  have 


R!  +  2  R2,  it  will  be  observed,  equals  P,  as  should  be. 

The  conception  of  a  beam  tipping,  as  in  the  last  two  Arts.,  is  due  to  Clem- 
ens Herschel  (Continuous,  Revolving  Drawbridges,  Boston,  1875),  and  the 
above  formulae  were  first  deduced  by  him  in  the  above  work. 


LITERATURE   UPON   THE    CONTINUOUS    GIEDEE.  247 


LITERATURE   UPON  THE   CONTINUOUS   GIRDER. 

"We  give  below,  for  the  benefit  of  students  and  those  interested  in  the 
subject,  a  list,  chronologically  arranged,  of  works  upon  the  continuous 
girder.  A  glance  at  this  list  will  convince  the  reader  as  to  the  thorough- 
ness with  which  the  problem  has  been  treated. 

1.  REBHANN. — "Theorie   der  Holz-und  Eisenconstructionen."      Wien, 
1856. — [Treats  the  continuous  girder  of  constant  cross-section  and  equal 
spans  according    to   the  old  method ;    first  determining  the  reactions  at 
the  supports.     A  load  in  any  single  span  only  is  considered,  either  total 
uniformly  distributed,  or  concentrated  and  acting  at  the  centre.] 

2.  KOPKE. — "  Ueber  die  Dimensionen   von  Balkenlagen,  besonders  in 
Lagerhausern."     Zeitschr.  des  Hannov.  Arch.  u.  Ing.  Ver.,  1856.      [The 
simple  and  continuous  girder.     Attention  is  here  first  called  to  the  advan- 
tage gained  from  sinking  the  supports.] 

3.  SCHEFFLER. — "  Theorie   der   Gewolbe,   Futtermauern  und  Eisernen 
Bracken."     Braunschweig,  1857.     [Continuous  girder  with  total  uniformly 
distributed  load,  and  invariable   concentrated  loading.      Advantage  of 
sinking  the  supports.] 

4.  CLAPEYRON. — Calcul  d'une  poutre  elastique  reposant  librement  gui- 
des appuis  inegalement  especes."     Comptes  rend  us,  1857.     [Here,  for  the 
first  time,  the  well-known  Clapeyronian  method  is  developed,  by  which  a 
series  of  equations  between  the  moments  at  the  supports  is  first  obtained. 
Application  to  total  distributed  loads,  but  varying  in  different  spans.] 

5.  MOLLINOS  ET  PRONIER. — "  Traite  theoretique  et  practique  de  la  con- 
struction des  ponts  metalliques."     Paris,  1857.      [Treatment  of  the  con- 
tinuous girder  of  constant  cross-section,  according  to  Clapeyron.] 

6.  GRASHOF. — "Ueber  die  relative  Festigkeit  mit  Riicksicht  auf  deren 
moglichste  Vergrosserung  durch  angemessene  Unterstiitzung  und  Einmau- 
erung  der  Trager  bei  constantern  Querschnitte."      Zeitschr.  des  Deutsch. 
Ing.  Ver.,  1857,  1858,  1859. 

7.  MOHR. — "  Beitrag  zur  Theorie  der  Holz-und  Eisenconstructionen." 
Zeitschr.  des  Hannov.  Arch.  u.  Ing.  Ver.,  1860.     [Theory  of  continuous 
girder,  with  reference  to  relative  height  of  supports.     Application  to  gird- 
ers of  two  and  three  spans.     Best  sinking  of  supports  for  constant  cross- 
section.     Disadvantage  of  accidental  changes  of  height  of  supports.     In- 
fluence of  breadth  of  piers.] 

8.  II. — "  Continuirliche  Briickentrager^"     Bornemann's  Civil-Ingenieur, 
1860.     [Continuous  girder  of  constant  cross-section  of  three  spans.     Best 
ratio  of  spans,  and  sinking  of  supports.] 

9.  WINKLER. — "Beitrage  zur  Theorie    der    continuirlichen    Briicken- 
triiger."      Civil-Ingenieur,    1862.      [General  Theory.       Determination  of 
methods  of   loading  causing  maximum  strains;    and,  for  the  first  time, 
general  rules  for  the  same  given.    Best  ratio  of  end  spans.] 


248  LITERATURE    UPON    THE   CONTINUOUS    GIRDER. 

10.  BUESSE. — "Cours  mecanique  applique*e  professe  a"  Fecole  imperiale 
des  ponts  et  chaussees."     Seconde  Partie.     Paris,  1862.     [Analytical  treat- 
ment of  the  continuous  girder  of  constant  cross-section.     The  transverse 
forces  are  not  considered.     The  exact  determination  of  the  most  dangerous 
methods  of  loading,  with  reference  to  the  moments  in  the  neighborhood  of 
the  supports,  is  also  wanting.] 

11.  ALBARET. — "Etude  des  ponts  metalliques  a  poutres  droits  reposant 
sur  plus  de  deux  appuis."     Ann.  des  ponts  et  chaussees,  1866.     [Continu- 
ous girder  of  constant  cross-section,  treated  after  Clapeyron.] 

12.  RENAUDOT. — "Memoire  sur  le  calcul  et  le  control e  de  la  resistance 
des  poutres  droites  a"  plusiers  travees."     Ann.  des  ponts  et  chaussees,  1866. 
[Continuous  girder,  treated  according  to  Clapeyron.] 

13.  CULMANN. — "Die  Graphische  Statik."      Zurich,  1866.     [Graphical 
treatment  of  simple  and  continuous  girder  of  constant  and  variable  cross- 
section.     Moments  at  the  supports  are  determined  analytically.] 

14.  H.  SCHMIDT.—"  Ueber  die  Bestimmung  der  ausseren  auf  em  Briick- 
ensystem  wirkenden  Krafte."     Forster's  Bauz.,  1866.     [Data  for  the  amount 
of  live  load  for  Railroad  and  Way  Bridges.     Determination  of  the  equiva- 
lent uniformly  distributed  load.     Data  for  dead  weight  and  wind  pres- 
sure. ] 

15.  GRASHOF. — "Die  Festigkeits  Lehre."      1866.      [General  analytical 
treatment  of  the  girder  without  special  reference  to  bridges.     Continuous 
girder  of  uniform  strength.] 

16.  WINKLER. — "  Die  Lehre  von  der  Elasticitaet  mid  Festigkeit."     Prag., 
1867. — [General  analytical  theory  of  the  continuous  girder  of  constant  and 
variable  cross-section.     Application  to  total  uniformly  distributed  loading. 
Influence  of  difference  of  height  of  supports.] 

17.  FRANKEL. — "Ueber  die  ungiinstigste   Stellung  eines   Systems  von 
Einzellasten  auf  Tragern  iiber  eine  und  iiber  zwei  Oeffnungen,  speciell  auf 
Tragern  von  Drehscheiben."     Bornemann's  Civil-Ingenieur,  1868. 

18.  MOHR. — "Beitrag  zur  Theorie  der  Holz-  und  Eisenconstructionen." 
Zeitschr.  des  Hannov.  Arch.  u.  Ing.  Vcr.,  1868.     [Here,  for  the  first  time, 
the  elastic  line  is  regarded  as  an  equilibrium   curve,  and  the  graphical 
treatment  of  the  continuous  girder  founded.] 

19.  H.  SCHMIDT. — "  Betrachtungen  iiber  Briickentrager,  welche  auf  zAvei 
und  mehr  Stiitzpunkte  frei  aufliegen,  sowie  iiber  den  Einfluss  der  unglei- 
chen  Hohenlage  der  Stiitzpunkte."     Forster's  Bauz.,  1868. 

20.  COLLIGNON. — "  Cours  de  mgcanique  applique*e  aux  constructions." 
Paris,  1869.     [Continuous  girder  of  constant  cross-section   and  uniform 
load.] 

21.  LAISSLE  and  SCHUBLER. — "Der  Bau  der  Briickentrager  init  beson- 
derer  Riicksicht  auf  Eisenconstructionen."      III.  Ami.,  I.  Theil.      Stutt- 
gart, 1869.     [Treatment  of  the  continuous  girder,  according  to  Clapeyron.] 

22.  LEYGUE. — "  Etude  sur  les  surcharges  a  conside"rer  dans  les  calculs 
des  tabliers  metallique  d'appr§s  les  conditions  generales  d'exploitation 
des  chemins  de  fer."     Paris,  1871. 

23.  LIPPICH. — "  Theorie  des  continuirlichen   Triigers  constanten  Quer- 
sehnittes.     Elementare  Darstellung  der  von  Clapeyron  und  Mohr  begriin- 


LITEKATUKE    UI>ON    THE    CONTINUOUS    GIKDER.  249 

deten  Analytischen  und  Graphischen  Methoden  und  ihres  Zusammen- 
hanges."  Forster's  Bauz.,  1871,  also  separate  reprint.  [The  geometrical 
constructions  are  deduced  from  the  analytical  formulae.] 

24.  SEEFEHLNER,  G. — "  A  tobbnyngpontu  vasr&cstartokrdl — A  magyar 
m6rrok  es  Spitesz— egylet  kozlonye,"  1871  [Hungarian]. 

25.  HITTER,  W. — "  Die  elastische  Linie  und  ihre  Anwendung  auf  den 
continuirlichen  Balken.     Ein  Beitrag   zur  graphischen  Statik."     Zurich. 
1871.     [This  and  the  preceding  work  treat  the  continuous  girder  after  the 
Culmann-Mohr  method.] 

26.  OTT.— "  Vortrage   uber    Baumechanik,"    II.  Theil.      Prag,    1872. 
[Analytical  determination  of  the  shearing  forces  and  moments  for  the  sim- 
ple and  continuous  girder  of  constant  cross-section  and  level  supports.] 

27.  WEYRAUCH,  J.  I. — "  Allgemeine  Theorie  und  Berechnung  der  con- 
tinuirlichen und  einfachen  Trager."     Leipzig,  1873.     [A  work  well  deserv- 
ing to  close  the  list.     Gives  the  general  theory  for  constant  and  variable 
cross-section  for  any  number  of  spans  from  1  to  oo,  and  for  all  kinds  of 
regular  or  irregularly  distributed  and  concentrated  loads.     The  formulae 
are  general,  and  for  given  loading  free  from  integrals.     Difference  of  level 
of  supports ;  most  unfavorable  position  of  load ;  exact  theory  of  the  fixed 
and  movable  inflexion  and  influence  points,  etc.     Examples  illustrating 
use  of  formulae,  and  complete  calculations  of  girders.] 

This  last  work  leaves  but  little  to  be  desired  in  thoroughness  and  com- 
prehensiveness. 

It  will  be  observed  that  England  and  America  have  contributed  but  lit- 
tle to  the  literature  of  the  subject.  Indeed  the  standard  works  of  both 
countries  show  scarcely  a  trace  of  the  influence  of  the  labors  of  French  and 
German  mathematicians  in  this  field.  The  only  works  which,  so  far  as  we 
are  aware,  can  be  mentioned  in  this  connection  are  as  follows : 

RANKINE,  W.  J.  M. — "  Civil  Engineering."  1870.  [Very  brief  and  in- 
complete.] 

HUMBER. — "  Strains  in  Girders."  American  Ed.  New  York  :  Van 
Nostrand.  1870.  [Graphical  constructions,  holding  good  only  under  the 
supposition  that  the  end  spans  are  so  proportioned  that  the  girder  may  be 
considered  as  fixed  at  the  intermediate  supports,  for  full  load.] 

STONEY,  B.— "  Theory  of  Strains."  London,  1873.  [Very  brief  notice 
of  the  subject.  Points  of  inflection  are  found  for  full  load,  and  the  flanges 
then  cut  at  these  points.] 

HEPPEL,  J.  M.— Phil.  M^.  (London),  Vol.  40,  p.  446. 

Also  Minutes  of  the  Proceed,  of  the  Inst.  Civ.  Eng.  [Excellent  papers, 
which  might  well  have  been  followed  up.] 

In  the  latter  publication  also : 

BELL,  W.— Vol.  32,  p.  171. 

STONEY,  E.  W.— Vol.  29,  p.  382. 

BARTON,  JAMES.— Vol.  14,  p.  443. 

In  American  literature : 

FRIZELL. — "Theory  of  Continuous  Beams."  Jour.  Frank.  Inst.,  1872. 
[Davelopment  of  the  subject  according  to  Scheffler.  See  3.] 


250  LITEEATUEE    UPON   THE    CONTINUOUS    GIEDEE. 

GREENE,  CHAS.  E. — "Graphical  Method  for  the  Analysis  of  Bridge 
Trusses."  Van  Nostrand.  1875.  [Force  and  equilibrium  polygons  are 
used,  but  the  moments  at  the  supports  are  found  by  an  original  method  of 
approximation,  or  balancing  of  moment  areas. ,] 

HERSCHEL,  CLEMENS. — "  Continuous,  Revolving  Drawbridges."  Boston, 
1875.  [The  formulae  of  Wey ranch  are  made  use  of.  The  case  of  "  second- 
ary central  span "  is  for  the  first  time  investigated,  and  the  appropriate 
formulas  given.  The  fact  that  the  live  load  reactions  for  supports  out  of 
level  are  unchanged,  provided  the  dead  load  reactions  are  zero,  is  also  for 
the  first  time  clearly  stated.  The  draw  span  is  thoroughly  treated,  and  the 
idea  of  weighing  off  the  reactions  at  the  piers  of  a  continuous  girder  sug- 
gested.] 

MERRIMAN,  MANSFIELD. — "  Upon  the  Moments  and  Reactions  of  the 
Continuous  Girder" — Journal  of  the  Franklin  Inst..for  March  and  April, 
1875;  Van  Nostrand's  Eng.  Mag.,  July,  1875;  also  the  London  Phil. 
Mag.,  Sept.,  1875,  as  well  as  the  formulae  contained  in  Chapter  XII.  of  this 
work.  [By  the  aid  of  the  properties  of  the  Clapeyronian  numbers,  Mr. 
Merriman  has  deduced  new  and  general  formulae  eminently  suited  for 
practical  use.  Also  relations  are  deduced  from  which  tables  for  moments 
and  reactions  may  be  drawn  up  to  any  desired  extent  by  simple  additions 
and  subtractions,  independently  of  the  general  formulae.  (See  Chap.  XII.) 
The  simple  girder  appears  as  a  special  case  of  the  continuous  girder.  The 
formulae  are,  in  respect  to  simplicity  and  ease  of  application,  superior  to 
any  heretofore  given.] 


CHAP.  XIV.]  THE   BEACED  AECH.  251 


PART    III. 

APPLICATION  OF  THE  GRAPHICAL  METHOD  TO  THE  ARCH. 


CHAPTEK  XIV. 

THE   BKACED    AECH. 

157.  Different  kinds  of  Braced  Arches. — Just  as  in  gird- 
ers, we  may  distinguish  between  the  solid  beam,  or  "  plate  gird- 
er," and  the  open  work,  or  framed  girder ;  so,  regarding  the  arch 
as  a  bent  beam,  we  may  distinguish  the  braced  arch  and  the 
solid  arch,  or  arch  proper.     The  strains  in  the  various  pieces 
composing  the  braced  arch  may  be  easily  found  by  the  method 
of  Arts.  8-15,  or  by  calculation  by  the  method  of  moments  of 
Art.  14  for  any  loading,  if  only  all  the  outer  forces  acting 
upon  the  arch  are  known :  that  is,  so  soon  as,  in  addition  to  the 
load,  we  know  also  the  reactions  at  the  abutments,  or  the  hori- 
zontal thrust  and  vertical  reactions  at  the  points  of  support,  and 
the  moments,  if  any,  which  exist  at  these  points. 

We  may  distinguish  three  classes  of  braced  arches  :  viz.,  1st. 
Arch  hinged  at  both  crown  and  springing ;  2d.  Arch  hinged 
at  spring  line  only — continuous  at  crown ;  3d.  Arch  continu- 
ous at  crown  &&& fixed  at  abutments. 

158.  Arch  hinged  at  both  Crown  and  Abutments. — This 
form  of  construction  [PL  23,  Fig.  90],  owing  to  the  hinges  at 
crown  and  abutments,  affords  for  live  load  but  little  of    the 
advantage  of  a  true  arch.     It  is,  in  fact,  an  arch  only  in  form, 
but  in  principle  is  more  nearly  analogous  to  a  simple  triangular 
truss  of  two  rafters,  these  rafters  being  curved  and  braced  ;  the 
thrust  being  taken  by  the  abutments,  instead  of  resisted  by  a 
tie  line  A  B. 

The  case  presents  no  especial  difficulty,  and  may  be  easily 
calculated  or  diagramed,  provided  that  not  more  than  two 
pieces,  the  strains  in  which  are  unknown,  meet  at  any  apex. 
Thus,  in  PI.  23,  Fig.  90,  the  resultant  at  the  abutment  due  to 
any  weight  P  being  known,  it  may  be  directly  resolved  into  tnc 


252  THE   BRACED   AKCH.  [CHAP.  XIV. 

two  pieces  which  meet  there.  The  strains  in  these  two  pieces 
being  thus  found,  those  in  two  others  in  equilibrium  with  each 
of  them  may  be  obtained.  In  Art.  13  we  have  already  illus- 
trated the  method  of  procedure  for  such  a  case,  as  also  the 
method  of  finding  graphically  the  resultant  at  crown  and  abut- 
ments due  to  any  position  of  the  weight. 

Thus  the  resultant  at  the  crown  for  the  unloaded  half  must, 
for  equilibrium,  pass  through  the  hinge  at  B  also.  Its  direc- 
tion is  thus  constant  for  all  positions  of  P  upon  the  other  half. 
The  resultant  for  the  other  half  must  then  pass  through  a  and 
the  hinge  at  A  (Fig.  90). 

We  have  then  simply  to  draw  a  B,  prolong  P  to  intersection 
a,  and  draw  a  A.  A  a  and  B  a  are  the  directions  of  the  resul- 
tant at  A  and  B,  and  by  resolving  P  along  these  lines,  we  may 
find  the  vertical  reaction  V  =  d  l>  and  the  horizontal  thrust 
H  =  cb. 

We  can  thus  easily  find  the  reactions  at  the  abutments  in 
intensity  and  direction,  and  following  these  reactions  through 
the  structure,  as  illustrated  in  Arts.  8-13,  Chap.  I.,  can  deter- 
mine the  strains  upon  all  the  pieces  for  any  position  of  the 
weight.  A  tabulation  of  the  strains  for  each  weight  will  then 
give  us  the  strains  for  uniform  load  as  well  as  live  load,  as  al- 
ready explained  in  the  preceding  chapter,  Art.  156. 

There  must  be  only  two  pieces  meeting  at  the  abutments. 
Thus  the  pieces  in  Fig.  90,  represented  by  broken  lines,  can 
serve  only  to  support  a  superstructure,  or  transmit  load  to  the 
arch,  and  have  no  influence  upon  the  strains  in  the  other  pieces. 

If  the  span  A  B  =  2  «,  the  rise  of  the  arch  is  A,  and  the  dis- 
tance of  the  weight  P  from  the  crown  is  x,  positive  to  the  left ; 
then  taking  moments  about  the  end  B,  we  have 

2  V0  =  P  (a  +  a),  or  V  =  P  (»  +  ^- 

A  Cb 

Similarly,  taking  moments  about  the  crown, 

„,  n  V  a—  Px      P  (a  —  x) 

—  v  #  +  H  A  =  —  P  #,      or   H= =—  -  —  — ^— ^ — '-. 

fi  2t  fi 

The  same  formulae  apply  for  a  weight  upon  the  other  half,  for 
V  and*  H  at  the  other  end. 

The  values  of  V  and  H  can  easily  be  found  from  these  for- 
mulae, and  the  strains  then  calculated  by  moments,  thus  check- 
ing the  diagrams.  If  these  reactions  are  found  for  the  given 


CIIAr.  XIV.]  THE   BRACED    ARCH.  253 

dimensions  of  the  centre  line,  we  may,  if  we  choose,  suppose 
the  depth  of  the  arch  to  vary  above  and  below  the  centre  line 
equally,  from  the  crown  to  ends.  The  lever  arms  of  the  pieces, 
and  hence  their  strains,  will  be  different,  but  V  and  H  are  the 
same  as  before.  Thus,  whatever  the  shape  of  the  arch,  we  can 
easily  find  the  strains  both  by  diagram  and  calculation.  If  we 
draw  a  line  through  A  and  the  hinge  at  crown,  we  may  easily 
prove  that  the  greatest  vertical  ordinate  between  this  line  and 
the  arch  is 


y  =  — 

O  /v 

where  r  is  the  radius. 

Now  if  the  depth  d  of  the  arch  is  made  greater  than  this  or- 
dinate,  it  may  be  shown  that  both  flanges  will  always  be  in 
compression.  This  condition  serves,  then,  to  determine  the 
proper  depth  of  circular  arch,  which  should  not  be  less  than 


It  is  unnecessary  to  give  here  an  example.*  The  method  is 
so  simple  that  the  reader  will  find  no  difficulty  in  applying  the 
principles  above  to  any  case.  He  will  do  well  to  calculate  or 
diagram  the  strains  in  an  arch  similar  to  that  shown  in  the  Fig. 
for  comparison  with  the  two  cases  which  follow. 

159.  Arcli  hinged  at  Abutments — continuous  at  Crown. 
— If  we  suppose  the  hinge  at  the  crown  removed — those  at  the 
abutments  being,  however,  retained — then,  for  any  position  of 
the  weight,  the  resultant  at  each  end  must  for  equilibrium 
pass,  as  before,  through  the  end  hinges.  In  the  preceding 
case,  a,  for  load  on  left  half,  was  always  to  be  found  at  intersec- 
tion of  weight  with  the  line  through  B  and  hinge  at  crown,  and 
was  therefore  fully  determined.  Now,  however,  &,  the  com- 
mon intersection  of  weight  and  resultant  abutment  pressures, 
has  a  different  position,  and  hence  the  resultants  and  horizon- 
tal and  vertical  reactions  are  different. 

If  we  can  find  or  know  the  locus  or  curve  in  which  this  point 
a  must  always  lie,  we  can  easily  find,  as  before,  the  resultants 
or  reactions  by  simply  prolonging  the  line  of  direction  of  the 
weight  till  it  meets  this  locus,  and  then  drawing  from  the  point 

*  See  Note  to  this  Chap,  in  Appendix. 


254: 


THE   BRACED   ARCH. 


[CHAP.  xiv. 


of  intersection  lines  to  A  and  B,  and  resolving  P  in  these  direc- 
tions. 

The  equation  of  this  locus  can  be  found  analytically  without 
much  difficulty. 

1st.  PARABOLIC  ARC. — Thus,  for  &  parabolic  arc,  we  have* 

32  a?  h 
y  ~  5  (5  ^  -  ar8)' 

Where  [PL  23,  Fig.  91]  a  is  the  half  span,  and  h  the  rise  of 
the  arc ;  x  the  distance  of  the  weight  from  the  crown,  and  y 
the  ordinate  N  dot  the  locus  cd  eiJc. 

For  a  given  arc,  then — that  is,  a  and  h  given — we  have  only  to 
substitute  different  values  for  a?,  as  x  —  0,  0.1,  0.2,  etc.,  of  the 
span,  and  we  can  easily  find  the  corresponding  ordinates  y,  and 
thus  construct  the  locus  cdeiJc.  It  is  then  easy  to  find  the 
reactions  at  A  and  B  for  any  position  of  P,  as  above  indicated. 

The  vertical  reaction  at  the  abutment  may  also  be  easily 
found  by  moments — thus, 

Vt  x  2  a  =  P  (a  +  a?),   or  Vt  =  —  (a  +  x). 

2i  a 


The  horizontal  thrust  is 
5 


(5  a9  -  a?)  (a2  -  a?)  » 
a3  A 


These  values,  though  not  needed  for  the  construction  above, 
may  be  of  use,  and  are  therefore  given.  In  the  following 
tables  we  give  the  values  of  H  and  y  for  different  values  of  x : 


X 

H 

y 

x 

H 

y 

0 

0.3906 

1.280 

0.5 

0.2783 

1.347 

0.1 

0.3859 

1.283 

0.6 

0.2320 

1.379 

0.2 

0.3706 

1.290 

0.7 

0.1797 

1.415 

0.3 

0.3490 

1.304 

0.8 

0.1226 

1.468 

0.4 

0.3176 

1.322 

0.9 

0.0622 

1.527 

0.5 

0.2783 

1.347 

1.0 

0 

1.620 

.a 

£ 

.h 

.« 

ft 

.A 

*  For  the  demonstration  of  the  analytical  results  made  use  of  in  this  chap- 
ter, we  refer  the  reader  to  Die  Lehre  von  der  Elasticitat  und  Festigkeit,  by  E. 
Winkler.  Prag,  1867.  See  also  the  Supplement  to  this  chapter. 


CHAP.  XIV.]  THE  BRACED  ARCH.  255 

From  the  table,  a  and  h  and  P  being  known,  H  and  y  can  be 
found  for  the  successive  positions  of  P  at  0.1,  0.2,  etc.,  of  #,  or 

the  half  span,  by  multiplying  P  —  by  the  tabular  number  for 

H,  and  h  by  the  tabular  number  for  y. 

2d.  CIRCULAR  ARC. — For  a  circular  arc  we  have  for  the  equa- 
tion of  the  locus  cdeik  [Fig.  91], 


where  K  =  -r— 3,  I  being  the  moment  of  inertia  of  the  constant 

cross-section,  A  its  area,  and  r  the  radius  of  the  circle :   also 
where 

(sin2  a— sin2  j3)  (a— 3  sin  a  cos  a  +  2  a  cos2  a) 

sin  a  [sin2  a— sin'2  /3  +  2  cos  a  (cos  /3— cos  a)  — 2  cos  a  (a  sin  a— £  sin  0)]' 

a  being  the  angle  subtended  at  centre  by  the  half  span,  and 
.  __  2  cos  a  (a  sin  a  —  /3  sin  /3) 

B  = 


sin2  a  -  sin2  /3  +  2  cos  a  (cos  @  —  cos  a  —  a  sin  a  H-  /3  sin 
2  a  cos2  a 


2  (a  —  3  sin  a  cos  a  +  2  a  cos2  a)' 
or,  approximately, 

24  6  aa 


5  A2 

=  IT^4  =  64  P 
where  $  is  the  angle  from  crown  to  weight. 

—  —  the  square  of  the  radius  of  gyration,  or,  approximately, 
the  square  of  the  half  depth,  hence  K  =  -_  approximately. 


256  THE   BRACED    ARCH.  [CHAP.  XIV. 

For  the  exact  values  of  A  and  B,  we  have  the  following  table : 


fi 

a=  0 

a=10° 

a=20° 

a-30- 

a-=40° 

a=50° 

a=W 

a=90° 

0 

1.20 

1.19 

1.17 

1.14 

1.08 

1.00 

0.88 

0 

0.2 

1.21 

1.20 

1.18 

1.15 

1.10 

1.01 

0.90 

0 

A 

0.4 
0.6 

1.24 
1.29 

1.24 
1.29 

1.21 
1.27 

1.18 
1.24 

1.13 
1.20 

1.05 
1.13 

0.94 
1.02 

0 
0 

- 

0.8 

1.88 

1.38 

1.36 

1.34 

1.30 

1.24 

1.18 

0 

1.0 

1.50 

1.50 

1.49 

1.47 

1.45 

1.41 

1.36 

0 

B 

a 

0.234 

0.233 

0.221 

0.203 

0.178    0.146 

1 

0.107 

0 

a4 

For  the  values  of  yQ  we  have  the  following  table 


£ 

a=0 

•=10- 

a=20° 

a=30° 

«=«• 

a=50° 

.=60- 

a=W 

0 

1.280 

1.282 

1.288 

1.300 

1.316 

1.340 

1.375 

1.571 

0.2 

1.290 

1.292 

1.298 

1.309 

1.327 

1.348 

1.380 

1.571 

0.4 

1.322 

1.324 

1.320 

1.340 

1.354 

1.374 

1.403 

1.571 

0.6 

1.379 

1.380 

1.385 

1.393 

1.405 

1.421 

1.443 

1.571 

0.8 

1.468 

1.4G9 

1.471 

1.476 

1.483 

1.490 

1.504 

1.571 

1.0 

1.600 

1.600 

1.599 

1.597 

1.594 

1.591 

1.588 

1.571 

.a 

.h 

It  will  be  seen  that  for  the  semi-circle  the  locus  is  a  straight 
line,  for  which  y  =  ^irr  =  1.5708r.  Thus,  for  any  given  case 
—that  is,  I,  A  and  r  given — we  can  easily  calculate  K.  Then 
from  our  tables,  for  given  value  of  a,  we  can  find  A,  B  and  yQ  for 
values  of  ft  of  0,  -j^ths,  T\ths,  etc.,  of  a.  These  values  inserted 
in  the  equation  for  y  above,  will  enable  us  to  plot  the  curve  or 
locus  cdeik)  which  being  once  known,  the  rest  is  easy.  We 
have  thus  by  a  union  of  analytical  results  with  our  graphical 
method  a  very  easy  and  practical  solution  of  this  important 
case.  We  may,  jf  we  choose,  only  use  our  method  to  determine 
the  horizontal  thrust  and  vertical  reaction  as  shown  by  the  Fig. 
91,  and  then  calculate  the  strains  by  the  method  of  moments. 
The  availability  and  ease  of  the  method  here  given,  as  com- 
pared with  calculation,  will  be  seen  from  a  consideration  of  the 


CHA.P,  XIV.]  THE   BRACED    ARCH.  257 

analytical  formulae  for  the  horizontal  thrust  and  vertical  reac- 
tion at  A.  Thus,  for  the  vertical  reaction,  we  have,  as  before, 
simply 


For  the  horizontal  thrust,  however,  we  have  the  following 
very  clumsy  formula  : 

sin*  a  —  sin2  p  +  2  cos  a  (cos  {I—  cos  a)  —  2  (l-f-ic)cosa  (a  sin  a—  £sin/9) 
2  [a  —  3  sin  a  cos  a  +  2  (1  +  *)  a  cos2  a] 

For  the  semi-circle,  this  reduces  to 


K  being,  as  before,  =  -r— ^ ;    where  A  is  the   area  and  I  the 

moment  of  inertia  of  the  cross -section,  r  the  radius  of  the  arch, 
and  the  angles  a  and  /3,  as  represented  in  Fi^.  91,  viz.,  the 
angle  of  the  half  span,  and  the  angle  to  the  load,  subtended  by 
x.  The  first  of  the  above  formulae  is  sufficiently  simple,  and 
by  it  we  may  check  the  accuracy  of  our  construction.  Thus 
having  plotted  the  curve  cdeiJc  by  the  aid  of  our  expression 
for  y  and  the  tables  above  for  any  position  of  P  required,  we 
draw  d  A  d  B,  and  resolve  P  along  these  lines,  thus  finding  V 
and  H  [Fig.  91].  We  can  then  calculate  V  from  the  formulae 

p 

above,  viz.,  V  =  —  (a  +  x).     If  this  calculated  value  agrees 

2  Ob 

with  that  found  by  diagram,  we  may  have  confidence  that  the 
curve  is  properly  plotted,  and  hence  that  the  value  of  H  is  also 
correct.  Thus,  with  very  little  calculation  and  great  ease, 
rapidity  and  accuracy,  we  can  find  the  reactions  at  the  end  A 
for  any  given  position  of  P  in  any  given  case.  These  reactions 
once  known,  we  can  easily  find  the  strains  either  by  diagram, 
as  illustrated  in  Chap.  1.,  or  by  calculation  by  the  method  of 
moments  of  Art.  14. 

16O.  Arch  fixed  at  Abutments— continuous  at  Crown. 
This  is  by  far  the  most  important  case  of  braced  arch,  as  by  the 
continuity  of  the  crown  and  fixity  of  ends  we  obtain  all  the 
advantage  possible  due  to  the  combined  strength  and  elasticity 
of  the  arch.  It  is  also  the  most  difficult  case  of  solution,  as  the 

formulae  obtained  by  a  mathematical  investigation  are  complex, 
17 


258  THE   BE  AGED    ARCH.  [CHAP.  XJV. 

and  give  rise  to  tedious  and  laborious  computations  in  practice. 
A  method  combining  simple  analytical  results  with  graphical 
construction  similar  to  the  preceding,  will,  however,  obviate 
these  difficulties,  and  bring  the  subject  fairly  within  the  reach 
of  the  practical  engineer. 

In  the  present  case,  as  before,  the  common  intersection  of  the 
weight  and  the  reactions  lies  in  a  curve,  the  equation  of  which 
may  be  found,  and  the  curve  itself  thus  plotted  for  any  given 
case. 

But  this  curve,  or  locus,  ILK  [PI.  24,  Fig.  92]  being  con- 
structed, in  order  to  find  the  directions  of  the  reactions  which 
now  no  longer  pass  through  the  ends  of  the  arc  A  and  B,  it  is 
necessary  to  find  and  construct  also  the  curve  enveloped  ~by  these 
reactions  for  every  position  of  P ;  that  is,  the  curve  to  which 
these  reactions  are  tangent.  If,  then,  these  two  curves  are  con- 
structed, we  have  only  to  draw  through  L  [Fig.  92]  lines  tan- 
gent to  this  enveloped  curve,  and  we  have  at  once  the  reactions 
in  proper  direction,  and  by  resolving  P  along  these  lines,  can 
easily  find  their  intensities,  and  therefore  V  and  H,  as  before. 

\st.  PARABOLIC  ARC. 

For  a  parabolic  arc  we  have  for  the  locus  ILK,  y  =  ^  h ; 
that  is,  the  locus  is  a  straight  line  at  \th  the  rise  of  the  arch 
above  the  crown  •  since  we  now  take  y  as  the  ordinate  to  the 
locus  measured  above  the  horizontal  tangent  at  the  crown.  The 
origin  is,  therefore,  at  the  crown  instead  of  at  the  centre  of  the 
half  span,  as  in  the  previous  case. 

For  the  second  curve,  or  curve  enveloped  by  the  reactions,  we 
have,*  taking  v  as  the  abscissa  and  w  as  the  ordiuate  of  any 

•   4.  n?-      oon  2  <#  (23  a2 +  20  ax  +  5  a?)  k 

point  [Fig.  92],  v  = ,     w  =  ^— , .  ,- f- , 

3  a  +  x '  15  (a  +  x)  (3  a  +  x) 

where,  as  before,  a  is  the  half  span,  h  the  rise,  and  x  the  dis- 
tance of  the  weight  from  crown.  For  x  =  0,  v  =  f  a,  and 
w  =  ff  h.  For  x  =  a,  v  =  £  a,  and  w  =  f  h.  For  x  =  —a, 
v  =  a,  and  w  =  —  oo.  Eliminating  x  from  both  equations,  we 

5  a2  —  5  av  +  2v*  t 

have  ^K  -  h. 

15  a  (a  —  v) 

Hence  the  curve  enveloped  by  the  reactions  is  on  each  side  an 

*  For  the  proof  of  all  the  expressions  assumed,  see  the  Supplement  to  this 
chapter. 


CHAP.  XIV.] 


THE   BEACED   ARCH. 


259 


hyperbola,  which  has  for  asymptotes  the  vertical  through  the 
abutment  and  a  straight  line  which  cuts  the  axis  of  symmetry 
of  the  arch  at  the  point  b  [PL  24,  Fig.  93],  ^  h  under  the  crown, 
the  tangent  at  the  crown  at  f  a  from  the  crown,  and  the  chord 
of  the  arc  at  —  6  a  from  the  centre.  The  centre  of  the  hyper- 
bola is  at  e,  yV  h  below  the  horizontal  through  the  crown.  The 
two  hyperbolas  osculate  at  the  point  -5-  a  vertically  below  the 
crown.  [See  Fig.  93.] 

As  an  aid  to  the  construction  of  these  hyperbolas,  we  give  the 
following  ^table : 


oc 

V 

w 

X 

V 

w 

1 

1.0000 

00 

0 

0.6667 

0.5111 

0.9 

0.9524 

2.7721 

0.1 

0.6452 

0.4897 

0.8 

0.9091 

1.5455 

0.2 

O.B249 

0.4722 

0.7 

0.8695 

1.1065 

0.3 

0.6061 

0.4577 

0.6 

0.8334 

0.9999 

0.4 

0.5882  . 

0.4463 

0.5 

0.8000 

0.7600 

0.5 

0.5714 

0.4349 

0.4 

0.7693 

0.6756 

0.6 

0.5555 

0.4258 

0.3 

0.7407 

0.6155 

0.7 

0.5405 

0.4160 

0.2 

0.7144 

0.5714 

0.8 

0.5263 

0.4102 

0.1 

0.6897 

0.5377 

0.9 

0.5128 

0.4053 

0 

0.6667 

0.5111 

1.0 

0.5000 

0.4000 

.a 

.a 

.h 

.a 

.a 

.h 

From  the  table  it  is  easy  to  construct  the  hyperbola  for  any 
given  case.  We  have,,  of  course,  a  perfectly  similar  hyperbola 
for  the  other  half,  its  centre  e  being  similarly  situated  with 
respect  to  the  crown,  to  the  right  of  c.  "We  have  then  simply 
to  draw  a  line  through  the  intersection  m  of  the  weight  P 
[Fig.  93]  with  the  line  i  k,  at  -g-  h  above  <?,  tangent  to  the  hyper- 
bola, and  we  have  at  once  the  direction  of  the  resultant.  This 
tangent  may  be  drawn  by  eye,  or  geometrically  constructed  if 


THE   BRACED    ARCH.  [CHAP.  XIV. 

desired.*  A  similar  tangent  to  the  hyperbola  on  the  other  side 
determines  the  direction  of  the  other  reaction.  We  can  then 
resolve  P  in  these  two  directions,  and  find  at  once  V  and  H. 
The  problem,  then,  so  far  as  a  parabolic  arc  is  concerned,  is 
sufficiently  simple  and  easy  of  solution.  We  have  only  to  draw 
a  straight  line  and  two  easily  constructed  curves.  The  formulae 
for  V  and  H  and  moment  at  crown  M0  are  for  this  case  also 
simple,  and  may  be  used  for  checking  our  results.  They  are  : 

TVr  -         IP  (a  ~  XY  (3  ^  ~  10  a  x  ~  5  ^ 
°~       *»  ~> 


H-  !*p-  V  - 

"J  - 


(a  ~ 


where,  as  before,  a  is  the  half  span,  h  the  rise,  and  x  the  dis- 
tance of  weight  P  from  crown.  A  negative  moment  always 
indicates  tension  in  lower  or  inner  flange. 

2d.  CIRCULAR  ARC. 

In  this  case  we  'have  for  the  locus  ILK  [Fig.  92],  for  small 
central  angles  a,  the  equation  : 


#,  A,  and  x  being  as  above,  and  <f  —  —r  —  the  square  of  radius 


of  gyration ;  A  being  the  area  and  I  the  moment  of  inertia  of 
cross- section. 

[Note. — In  all  the  cases  hitherto  considered,  or  which  we 
shall  consider  hereafter,  the  cross-section  is  assumed  constant.'] 

According  to  the  exact  formula,  which  is  too  complicated  to 
make  it  desirable  to  be  given  here,  we  have  the  following 

*  For  the  construction  of  a  tangent  to  a  conic  section,  see  Appendix,  Fig.  4. 


CHAP.  XIV.] 


THE  BRACED  ARCH. 
TABLE,  FOR  VALUE  OF  y. 


261 


0 

a  =  0° 

a  =  30° 

a  =  60° 

a  =90° 

0 

0.200 

0.211 

0.252 

0.329 

0.2 

0.200 

0.210 

0.246 

0.312 

0.4 

0.200 

0.307 

0.236 

0.280 

0.6 

0.200 

0.202 

0.217 

0.228 

0.8 

0.200 

0.197 

0.200 

0.158 

1.0 

0.200 

0.188 

0.151 

0.082 

a 

h 

Instead  of  determining  the  curves  enveloped  by  the  reac- 
tions, the  expressions  for  which  are  in  this  case  somewhat  com- 
plicated, it  will  be  found  preferable  to  find  the  distances  c±  c2 
of  the  intersections  <p  and  ty  of  the  reactions  [see  Fig.  92]  with 
the  verticals  through  the  centres  of  gravity  of  the  end  cross- 
sections.  For  small  central  angles  a,  we  have 


C,  — 


15  (a  +  x)  I  A  A2 

___2A_r 
15  (a  +  x)  I 

where  0,  A,  A  and  a?,  have  the  same  signification  as  above. 

Since  I  divided  by  A  equals  the  square  of  the  radius  of  gyra- 
tion =  ^,  we  have 

2A 


4- 


15  (a  +  a?) 


[ 


For  braced  arches  when  the  material  is  nearly  all  in  the 
flanges,  the  material  in  the  bracing  being  very  small,  we  may 
call  the  radius  of  gyration  half  the  depth  of  the  arch  measured 
upon  the  radius  from  centre  to  centre  of  flanges ;  or  represent- 
ing this  depth  by  d, 


262  THE    BRACED    AECH.  [CHAP.  XIV. 

2A          . 

—  5a?  — 


[A  negative  result  indicates  that  the  distance  is  to  be  laid  off 
below  the  centre  of  cross- section.]  These  formulae  are  easy  of 
application,  and  sufficiently  exact  for  arches  whose  rise  is  small 

compared  to  the  span  ;  when  — -  is,  say,  not  greater  than  -^. 

2  a> 

All  the  above  formulae  are  for  constant  cross-sections.  Exact 
formulae  for  variable  cross-section  give  results  but  little  less, 
and  are  much  more  complicated.  The  effect  of  using  the  above 
formulae  is  therefore,  merely,  to  increase  slightly  the  coefficient 
of  safety. 

161.  We  are  now  able  to  determine  readily  and  accurately 
the  strains  in  the  various  pieces  of  braced  arches  hinged  at 
crown  and  abutments,  and    hinged  at   abutments  only.     We 
have  only  to  construct  in  each  case  the  reactions  at  the  abut- 
ments, as  explained  in  Arts.  158  and  159,  Figs.  90  and  91,  and 
then,  by  the  method  already  detailed  in  Arts.  8-13,  we  can  fol- 
low these  reactions  through  the  structure,  and  thus  find  the 
strains  in  each  piece  due  to  every  position  of  the  load.     We 
may  also,  having  found  the  reactions  for  given    position   of 
weight,  calculate  the  strain  in  each  piece  by  moments. 

For  the  case  of  the  arch  continuous  at  the  crown  w&&  fixed  at 
the  abutments,  we  must  remember  that  we  have  also  a  moment 
at  each  end  tending  to  cause  either  tension  or  compression  in 
the  inner  flanges  according  as  it  is  negative  or  positive.  The 
case  is  precisely  analogous  to  the  continuous  girder,  or  girder 
fixed  at  ends.  As  in  that  case  [see  Fig.  77,  Art.  Ill]  the 
moment  at  one  end,  as  B,  was  the  product  of  H  into  the  vertical 
distance  B  D,  so  here  the  moment  at  A  (Figs.  92  and  93)  is  the 
product  of  H  into  ct,  found  by  the  formulae  above.  This 
moment  can,  then,  be  easily  found  when  cl  and  H  are  known. 
We  can  then  lay  it  off,  according  to  the  directions  of  Art.  125, 
for  "  passing  from  one  span  to  another  of  a  continuous  girder," 
and  thus  commence  our  diagram  of  strains;  or  we  can  cal- 
culate the  strains  by  the  method  of  moments. 

162.  Illustration  of  method  of  Solution. — As  an  illustra- 


CHAP.  XIV.]  THE    BRACED    AECH.  263 

tion,  take  a  portion  of  a  braced  arch,  as  represented  in  PI.  24-, 
Fig.  94.  We  have  first  to  plot  the  upper  curve  or  locus  of  m 
for  the  given  dimensions  of  the  centre  line  of  the  arch.  This 
curve  once  plotted,  then,  for  any  position  of  the  weight,  we  have 
only  to  prolong  P  to  m,  and  draw  a  line  from  m  to  the  end  of 
centre  line  if  the  arch  is  hinged  at  ends,  or  to  <£  at  a  distance 
<?!,  above  or  below  the  end  of  centre  line  if  the  arch  is  fixed  at 
ends ;  cl  being  easily  found  from  our  formulae  above.  In  sim- 
ilar manner,  we  draw  a  line  from  m  to  the  other  end,  or  C2. 
Now  these  two  lines  are  the  resultants  of  the  outer  forces  P, 
and  by  simply  resolving  P  in  these  directions,  we  have  at  once 
V  and  H,  while  the  moment  at  the  end  Mt  =  —  H  c1?  positive 
if  it  tends  to  cause  compression  in  lower  flange,  or  since  c±  is 
negative  down,  if  it  acts  below  the  end. 

We  can  now  easily  find  the  strain  in  any  flange,  as  D, 
whether  the  arch  vary  in  depth  or  not,  provided  only  it  is  sym- 
metrical with  respect  to  its  centre  line.  Thus  for  D,  take  the 
opposite  apex  a  as  the  centre  of  moments.  The  moment  of  H 
with  reference  to  #,  as  shown  in  the  Fig.,  tends  to  cause  tension 
in  D,  while  that  of  V  causes  compression.  We  have  then, 
representing  tension  by  minus, 

moment  of  V  —  moment  of  H 

strain  in  D  = = „  — , 

lever  arm  or  D 

all  with  reference  to  a.  If  the  result  is  minus,  it  indicates  thus 
tension,  if  plus,  compression  ;  if  it  is  zero,  the  two  moments  are 
equal,  and  at  «,  therefore,  no  moment  exists ;  hence  a  must  be 
a  point  of  inflection.  Note  that  H  and  V  must  be  taken  as  act- 
ing at  $,  Fig.  94  We  can  also  evidently  take  them  as  acting 
at  the  centre  of  the  end  cross-section,  if  we  take  into  account 
the  moment  H  c^ 

In  similar  manner,  for  C  we  take  5  as  centre  of  moments,  and 
then,  since  H  now  causes  compression  in  C  and  V  tension,  we 
have  for  V  and  H,  acting  at  <£, 

.     .    ^       moment  of  H  —  moment  of  V 

strain  in  C  =  -  — — . 

lever  arm  or  C 

For  V  and  H,  considered  as  acting  at  the  end  of  centre  line,  we 

„       moment  of  H  +  H^  —  moment  of  V 
have          C  =  -  — =—  ,  — . 

lever  arm  or 

taking  <\  without  regard  to  its  sign,  but  simply  to  the  kind  of 


264:  THE    BRACED    ARCH.  [CHAP.  XIV. 

strain  it  tends  to  cause  in  the  piece  in  question.  Properly, 
since  when  H  is  belo\v  ^  is  negative,  we  should  have  —  H  ^ 
for  moment  causing  compression  in  C. 

Thus  we  may  proceed  till  we  pass  P,  and  then  the  moment 
of  P,  with  its  proper  sign,  as  producing  tension  or  compression 
in  the  piece  in  question,  must  also  be  taken  into  account,  or  we 
may  instead  take  the  moments  of  V  and  H  at  the  other  end, 
that  is,  the  same  side  of  the  weight  as  the  piece  itself. 

The  diagonals  may  be  similarly  found  by  moments.  It  will, 
however,  be  best  to  determine  them  by  diagram,  one  of  the 
flanges  being  first  calculated  (in  this  case  the  first  upper  flange), 
as  explained  in  Art.  125.  They  may  also  be  calculated  from 
the  resultant  shear  at  any  apex.  Thus,  for  diagonal  3  find  the 
vertical  components  of  the  previously  determined  strains  in  D 
and  C.  These  vertical  components,  together  with  the  vertical 
component  of  the  strain  in  3,  must  for  equilibrium  be  equal  and 
opposite  to  the  total  shear  at  ft. 

Calling  this  shear  F,  and  a,  fi  and  y  the  inclinations  of  D 
and  3,  we  have  for  the  strain  in  3, 

S8  =  (F  —  Sl  sin  a  —  S2  sin  ft)  cos  7. 

If  either  of  the  vertical  components  of  the  strains  in  D  or  C 
acts  opposite  to  the  shear  F,  it  must,  of  course,  be  subtracted  ;  if 
in  the  same  direction,  added  to  F.  For  the  ready  determina- 
tion of  the  proper  signs,  see  Appendix,  Art.  16  (4). 

The  moment  H  ^  is  the  moment  at  the  fixed  end,  and  is  con- 
stant throughout  the  arch  for  any  one  position  of  the  load.  It 
causes  tension  in  outer  and  compression  in  inner  flanges,  pro- 
vided, as  in  the  Fig.,  <f>  fall  below  the  centre  of  the  end  section. 
This  moment  is  increased  (or  diminished  if  <j>  is  above)  by  the 
varying  moment  of  H  for  each  apex. 

The  above  method  of  determining  the  strains  in  the  braced 
arch,  though  not  strictly  graphical,  but  rather  a  combination  of 
analytical  and  graphical  methods,  offers  such  a  ready  solution 
of  this  important  and  difficult  case,  that  we  have  not  thought  it 
out  of  place  to  notice  it  somewhat  in  detail.  We  consider  it  by 
far  the  simplest  and  easiest  method  which  has  yet  appeared. 

163.  Analytical  Formulae  for  V  and  H. — A  comparison 
of  our  method  with  the  long  and  involved  analytical  expres- 
sions to  which  the  theory  of  flexure  conducts  us,  will  render  its 
advantages  still  more  apparent. 


CHAP.  XIV.]  THE   BRACED   AKCH.  265 

Thus,  for  a  load  of  w  per  unit  of  horizontal  length,  reaching 
from  left  end  to  a  point  whose  angle  from  vertical  through 
crown  is  ft  (Fig.  92),  a  being  the  angle  subtended  by  the  half 
span,  we  have  * 


H  =  &  sin  (B-Jci  +.k  +     sin 


where  R  is  radius  of  arch,  and 

7  .  2  sin  a       ,        sin  a  cos  a    .   sin2  a 

fc  =  a  -r  sm  a  cos  a  —  . ,     \  =  -f  — s— , 

a  a  2 

7        sin  a      7        sin  a  r  n  -, 

#2  =  — — ,     #3  =  — —      a  —  sin  a  cos  a  ,     and 

/z,  =  /3  +  2  yS  sin2/3  +  3  sin  ^Q  cos  j3. 

For  V  we  have 

v_    RF     cos  a  sin2  8        sin/3  K  3  cos  a— cos3  a       .     "I 

L.2  (a— sin  a  cos  a)         2         a— sin  a  cos  a      6  (a —sin a  cos  a)  J' 

„,      cosyS      8  sin  8      coss/3 
Where  -^  +  '__J yZt 

For  a  concentrated  load  P  for  any  point  [Fig.  92],  we  havef 


or,  more  correctly, 

__  _  p  a  —  /3  —  sin  a  cos  a  —  sin  /3  cos  /3  +  2  cos  a  sin  ft 
2  (a  —  sin  a  cos  a) 

For  the  semi-circle,  this  becomes 

v  _  p  TT  —  2  /3  —  2  sin  /3  cog  ft 


2  TT 
For  H  we  have 


H 


—  p  2  sip  «  [C03  3  —  cos  a  +  (1  +  «)  /3  sin  3]  —  (1  +  «)  a  (sin2  a  +  sin2  ff) 
2  [(1  +  K)  (t  (a  +  sin  a  cos  a)  —  2  sin2  a] 


*  Taken  from  Capt .  Bads'  Report  to  the  lUinois  and  St.  Louis  Bridge  Co. , 
May,  1868. 
f  Die  Lehre  von  der  Eiasticitdt  und  Festigkeit.    Winkler.    Prag.  1867. 


266  THE    BKACED    AKCH.  [CHAP.  XIV. 

where  K  =  — — g ;  I  being  the  moment  of  inertia,  and  A  area  of 

the  cross-section,  and  r  the  radius  of  circle. 

These  formulae,  it  will  be  observed,  involve  much  labor  in 
any  particular  case.  Where  the  number  of  weights  is  large, 
the  computation  is  tedious  in  the  extreme.  A  method  which 
shall  give  accurate  results  and  avoid  such  formulae  as  the  above 
is  certainly  very  desirable,  and  such  we  believe  to  be  the 
method  which  we  have  given. 

For  the  analytical  investigation  of  arches,  and  the  demon- 
stration of  the  formulae  for  the  curves  of  which  we  have  made 
use,  the  reader  may  consult  Die  Lehre  von  der  Elasticitaet  und 
Festigkeit,  by  Dr.  E.  WinJder,  to  which  we  have  already  re- 
ferred, and  which  contains  a  thorough  discussion  of  the  whole 
subject.  The  tables  which  we  have  given,  as  well  as  the  for- 
mulae for  £/,  G!  and  <?2,  will,  it  is  hoped,  give  the  method  here 
presented  a  practical  value,  and  render  the  solution  of  any  par- 
ticular case  easy  and  rapid. 

164.  For  a  solid  or  plate  girder  arch  of  given  cross-section, 
we  may  also  determine  the  proper  proportions  by  finding,  as 
above,  the  moment  M  of  the  exterior  forces  at  any  point. 

TI 

Then  M  =  _, 

t 

where  T  is  the  strain  per  unit  of  area  in  any  fibre  distant  t 
from. the  axis,  and  I  the  moment  of  inertia  of  the  cross-section. 
Thus,  for  a  rectangular  cross-section  I  =  ^  b  d3,  where  b  is  the 
breadth  and  d  the  depth. 

Hence  M  =  %  T  I  d\ 

if  we  take  t  =  -. 

The  strain,  then,  per  unit  of  outer  fibre  will  be 

T_6M 

"  bd2' 

The  safe  working  strain  should  not  exceed  for  iron  5  tons 
per  sq.  inch  for  tension  and  4  tons  for  compression,  and  there- 
fore d  being  assumed,  we  can  easily  proportion  b  so  as  to  sat- 
isfy this  condition. 

As  examples  of  braced  arches,  such  as  we  have  considered, 
viz.,  continuous  at  crown  and  fixed  at  abutments,  we  may  men- 


CHAP.  XIV.]  THE   BRACED   AKCH.  267 

tion  the  Bridge  over  the  Mississippi  River  at  St.  Louis,  ~by 
Copt.  Eads  /  one  over  the  Elbe  near  Hamburg  on  the  Paris- 
Hamburg  R.  R.,  in  which,  however,  the  outward  thrust  of  the 
arch  is  balanced  by  a  precisely  similar  inverted  braced  arch, 
or  suspension  system.  Thus  the  piers  have  to  support  a  verti- 
cal reaction  only,  and  the  necessity  of  large  and  expensive  abut- 
ments of  masonry  for  resisting  the  horizontal  thrust  is  obviated. 

The  strains  in  the  inverted  arch  of  this  character  are  found 
in  a  precisely  similar  manner.  The  only  difference  is  that  the 
reactions,  and  therefore  the  vertical  and  horizontal  components, 
act  now  in  a  direction  opposite  to  the  direction  for  the  upright 
arch,  and  the  strains,  though  the  same  in  amount,  are  of  re- 
verse character  in  each  piece. 

The  bridge  over  the  Rhine  at  Coblenz  is  an  illustration  of 
the  braced  arch  pivoted  at  the  abutments  only. 

Examples  of  the  solid  or  cast-iron  arch  of  all  kinds  are 
common. 

165.  Strains  due  to  Temperature.  —  In  the  first  class  of 
braced  arches,  viz.,  pivoted  at  both  abutments  and  crown,  there 
are  evidently  no  strains  due  to  changes  of  temperature.  The 
arch  can  accommodate  itself  to  any  change  of  length  by  rising 
at  the  crown  and  turning  at  the  abutments,  and  no  strains  are 
induced. 

We  represent  by  e  the  coefficient  of  linear  expansion  for  one 
degree  [about  0.000012  for  iron,  for  every  degree  centigrade], 
and  by  t,  the  temperature  above  or  below  the  mean  tempera- 
ture £0,  for  which  no  strain  exists. 

Then  for  arch  pivoted  at  abutments  only,  we  have  for  the  in- 
crease of  thrust,* 

__  2  El  et  sin  a 
~  7*  (a  —  3  sin  a  cos  a  +  2  a  cos2  a)  +  2  K  r*  a  cos3  a, 
where  E  is  modulus  of  elasticity,  I  moment  of  inertia  of  cross- 

section,  and  K  =  —  —  —  ;  A  being  area  of  cross-section,  r  radius  of 
arch,  a  the  angle  of  half  span,  or,  approximately, 


8  A  A2  +  15  I' 
where  h  is  the  rise  of  arch. 

*Lehre  von  der  Elasticitdt.     Winkler.     Also  Supplement  to  this  chapter, 
Art.  26. 


268  THE   BKACED    ARCH.  [CHAP.  XIV. 

„  For  the  moment  at  any  point,  then,  due  to  change  of  temper- 
ature, we  have 

M  =  H  r  (cos  /3  —  cos  a), 

ft  being  the  angle  from  vertical  to  that  point. 

This  moment,  if  positive,  causes  tension  in  outer  and  com- 
pression in  inner  flanges,  and  we  can,  as  before,  easily  find  the 
corresponding  strains  either  by  diagram  or  calculation. 

For  an  arch  fixed  at  ends  and  continuous  at  crown,  we  have 


TT    


2  Ele  t  (1  +  K)  a  sin  a 


[(1  +  K)  (a2  -4-  a  sin  a  cos  a)  —  2  sin  2a]  ' 

or,  approximately, 

45  E  I  e  t  45  E  I  A  e  t 

~~  r2  (a4  +  45  K)  ~  4=  A  A2  +  45  I' 

But  this  thrust  does  not  act  at  the  abutment,  since,  if  it  did, 
there  would  be  no  moment  there.  It  must  be  considered  as 
acting  at  a  distance  for  rise  of  temperature,  below  the  crown  of 

(1  +  K)  a  —  sin  a 
e'--=         (l  +  «)a     -*•'• 

or,  at  a  distance  above  the  end  abutment  of  h  —  eQ. 
Approximately,  we  have 

_  a2  +  6  K      _  (  A  a2  +  6  I)  h 
6  3Aa?       ' 

a  being  the  half  span. 

This  thrust  and  its  point  of  application  being  known,  we  can 
easily  find  the  moment,  and  hence  the  strains  at  any  point. 
We  see  that  the  horizontal  thrust  is  about  six  times  as  great  as 
for  the  case  of  an  arch  pivoted  at  both  ends. 

The  constant  moment  acting  at  the  abutment,  which  may  be 
considered  as  acting  at  every  point,  is 


_  /sin  a  \  „ 

MA  =  —  1  --  cos  a  t  H  r  ; 


it  acts  to  cause  compression  in  outer  and  tension  in  inner 
flanges  at  abutment.  If  we  find  this  moment,  we  can  then  con- 
sider H  as  acting  at  the  end,  and  then  we  have  for  the  moment 
at  any  point 


*  Capt.  Eads'  Report  to  the  Illinois  and  St.  Louis  Bridge  Co.,  May,  1868. 


CHAP.  XIV.]  THE  BRACED  ARCH.  269 

a  positive  result,  giving  tension,  a  negative,  compression  in  the 
outer  flanges. 

166.  Effects  of  Temperature. — We  are  now  able  to  solve 
accurately  and  thoroughly  any  class  of  braced  arch,  both  for 
variable  loading  and  changes  of  temperature,  and  here  the  fol- 
lowing remarks  upon  the  latter  subject  may  not  be  without 
interest.  We  quote  from  Culmann — Die  Graphische  Statik, 
p.  487  : 

"  The  question  arises  whether  the  fears  which  the  additional 
strains,  due  to  variations  of  temperature,  have  given  rise  to,  are 
well  founded.  Before  the  construction  of  the  Arcole  Bridge 
in  Paris  the  Engineer  Oudry  made  various  experiments  with 
a  rib  of  about  the  same  span  as  the  bridge  itself,  of  which  the 
following  seems  decisive  as  regards  the  present  question.  By 
driving  in  the  wedges  upon  which  the  rib  rested  above  and  be- 
low, he  could  raise  and  lower  the  crown  much  more  than  the 
distance  due  to  variation  of  temperature  without  diminishing 

its  supporting   capacity Oudry,  having  thus  assured 

himself  of  the  harmlessness  of  temperature  variations,  decided 
upon  broad  and  firm  bearing  surfaces. 

"  Interesting  observations  have  also  been  made  upon  the 
changes  of  form  of  the  cast-iron  arch  of  60  metres  span  over 
the  Rhone  at  Tarascon,  published  in  the  Annales  des  Ponts  et 
Chemins,  1854,  from  which,  however,  it  only  appeared  that  the 
changes  of  form  followed  slowly  the  temperature;  that  they 
were  less  than  the  received  coefficients  would  have  led  us  to 
expect,  and  were  nowhere  found  to  be  prejudicial. 

"  Since,  then,  this  question  appears  to  have  been  settled  more 
than  ten  years  ago,  may  we  not  fear  that  those  who  still  wish  to 
pivot  iron  may  some  day  seize  upon  the  idea  of  pivoting  stone 
arches  also ! 

"  Stone,  as  is  well  known,  expands  not  much  less  than  iron 
for  equal  changes  of  temperature,  and,  moreover,  its  modulus  of 
elasticity  is  much  less.  The  expanded  stone  arch  cannot  accom- 
modate itself  to  the  given  span,  therefore,  as  easily  as  the  iron 
arch,  and  it  would  then  be  clearly  more  advantageous  to  pivot 
the  stone  arch  !  As,  however,  such  a  clumsy  contrivance  would 
give  no  great  impression  of  stability,  we  feel  justified  in  recom- 
mending a  broad  and  solid  bearing  surface  for  all  arches." 

As  the  opinion  of  an  eminent  engineer,  the  above  may  not  be 
without  interest.  We  would  only  add  that,  according  to  the 


270  THE  BRACED  AKCH.  [CHAP.  XIV. 

accepted  formulae  for  temperature  strains  already  given,  the 
results  are  of  more  importance  than  the  above  remarks  would 
indicate.  As  will  be  seen  in  the  Appendix,  the  temperature 
strains  in  the  braced  arch,  fixed  at  ends  and  continuous  at 
crown,  are  very  considerable,  and,  if  the  formulae  are  accepted 
as  correct,  can  by  no  means  be  disregarded.  By  comparison  of 
our  numerical  results  for  the  three  cases  of  braced  arch  there 
given,  it  appears  that  the  one  hinged  at  crown,  and  springing, 
is  by  far  the  best  form  of  construction,  but  it  must  be  remem- 
bered that  a  different  proportion  of  span  to  height  and  depth 
may  considerably  affect  this  conclusion.  Upon  this  point  we 
refer  the  reader  to  Art.  28  of  the  Appendix. 

With  the  above,  we  conclude  our  discussion  of  braced  arches, 
or  arches  whose  weight  is  not  so  great  that  the  effect  of  the  live 
load  can  be  disregarded,  and  pass  on  to  the  stone  arch,  or  arch 
proper.* 

*  See  Appendix,  Art.  17,  for  a  practical  application  of  the  principles  of  this 
chapter. 


CHAP.  I.]  SUPPLEMENT   TO   CHAP.  XIV.  271 


SUPPLEMENT  TO  CHAPTEE  XIY. 


DEMONSTRATION    OF   ANALYTICAL   FORMULA   GIVEN   IN   TEXT. 

In  order  to  complete  our  discussion  of  the  braced  arch,  we  shall  now 
give  the  analytical  development  of  the  formulae  of  which  we  have  made 
use  in  the  preceding  chapter.  We  do  this  the  more  readily,  as  in  no  book 
of  easy  access  to  the  student  are  these  formulae  made  out.  In  the  work  of 
WinUer,  already  referred  to  in  the  text,  will  be  found  a  very  thorough 
discussion  of  the  subject.  We  shall  confine  ourselves  at  present  to  the 
case  of  a  single  concentrated  load. 


CHAPTER    I. 

GENERAL  CONSIDERATIONS  AND  FORMULAE  FOR  FLEXURE. 

1.  Fundamental  Equations — The  resultant  of  all  the  forces  act- 
ing upon  a  curved  piece  in  a  common  plane  may  be  decomposed  into  a 
force  normal  to  the  piece  N,  and  into  a  compressive  or  tensile  force  in  the 
direction  of  the  axis  or  of  the  tangent  to  the  axis  G ;  and  this  latter  force, 
if  taking  effect  above  or  below  the  axis,  acts  to  bend  the  piece,  and  gives 
rise  to  a  moment  as  well  as  to  a  compressive  or  tensile  force  G.  These 
forces  cause  corresponding  strains.  Thus,  if  P  is  the  tangential  strain  per 
unit  of  area  d  a,  then 


(1) 

while,  if  v  is  the  distance  of  any  fibre  from  the  axis, 

M (2) 


(a)  Coefficient  of  elasticity. 

Let  the  length  of  a  piece  be  s,  its  area  of  cross-section  A,  and,  as  above, 

G 

the  force  acting  upon  this  area  G.     Then  ^  will  be  the  force  per  unit  of 


area.     Let  the  displacement  [elongation  or  compression]  produced  by  this 
G 
A 


force      be  As;  the  sign  A  indicating  and  reading  "elongation.1"     Now 


272 


SUPPLEMENT    TO    CHAP.  XIV. 


[CHAP.  i. 


experiment  shows  that  within  narrow  limits,  i.e.,  within  the  elastic  limits, 
the  elongation  or  compression  is  directly  as  the  force  of  extension  or  com- 
pression. Supposing  that  this  held  true  always  for  all  values  of  A  «,  then, 

since  a  force  —  produces  a  displacement  A  s,  the  force  necessary  to  produce 


a  displacement  *,  will  be  — 

A  s 


E  = 


as  great.     Calling  this  force  E,  we  have 
Gs 


The  force  E,  then,  is  the  force  which  would  le  necessary  to  produce  a  dis- 
placement s  equal  to  the  original  length,  if  the  law  of  proportionality  of  the 
displacement  to  the  force  always  held  good  for  all  values  of  A  s.  This 


value 


E  = 


_ 

AA 


(3) 


we  call  the  coefficient  of  elasticity. 

From  (3)  we  easily  obtain  for  the  force  in  the  direction  of  the  tangent 
to.  the  axis 


=  EA 


A  s 


and  for  the  relative  displacement 

A  a 


G 

E  A 


(4) 
(5) 


(&)  Fibre  strain  P,  and  moment  M. 

As  seen  in  eq.  (1),  the  longitudinal  strain  upon  an  element  of  cross-sec- 
tion da  is  called  P.    In  a  curved  piece  conceive  two  cross-sections,  as 


CHAP.  I.]  SUPPLEMENT   TO    CHAP.  XIV,  273 

shown  in  the  Fig.,  as  A  O  B  D  perpendicular  to  the  axis  of  the  piece  m  n. 
Let  these  sections  be  infinitely  near;  then,  the  distance  J>a  upon  the  axis  is 
d  s.  Let  d  *v  be  the  length  of  any  fibre,  as  d  c,  before  the  change  of  form. 
Then,  after  deformation,  its  length  is  =  d  sv  +  A  d  sv.  But  if  d  (f>  is  the 
small  angle  between  the  normals,  dsv  =  ds  +  vd(f),  where  v  is  the  distance 
a  c  of  any  fibre  from  the  centre  of  gravity  of  the  cross-section.  After 
deformation,  d  s  becomes  ds  +  Ads,  and  d  <j>  becomes  d  $  +  Ad<f>,  and 
d  sv  becomes  d  sv  +  A  d  sv.  Hence  the  length  of  any  fibre  after  deforma- 
tion is  d  sv  +  A  d  sy  =  d  s  +  A  d  s  +  v  (d  <£  +  A  d  <£). 
Subtracting  this  from  the  eq.  for  d  sv  above,  we  have 


Therefore  the  ratio  of  the  change  of  length  to  the  original  length  of  fibre 

Ads 


is 


dsv 

If  r  is  the  radius  of  curvature,  then  rdd>  =  ds,  — —  =  — ;  hence 

d  s      r 

A  d  sv      r  A  d  s         A  d  0-i      r  ... 

-^=h^+Bir^i— (0) 


G 

From  eq.  (1)  we  have  the  strain  on  a  fibre  P  =  -r . 

From  eq.  (4),  G  =  E  A  — .     Hence  P  =  B  — .      In  the  present 


case 


—  is  given  by  (6)  ;  therefore 


ds  d  s 

Since  now  from  (1)  G  =    /  Pda,  we  have  from  (7) 

G__     Ads   r  da  Ad<f)   f*v  d  a 

E  =  r  ~d7J  7^o  +  '  ~~d~»J  r~TV 

Since  from  (2)  M  =    /  P  v  d  a,  we  have  again  from  (7) 

M_      Ads  rvda         Ad<f>   /V  d  a 
B  =    "~d~s~J  7T^+  "TT'J  7+~V 

But   /  -  —    is,   when  9  is  very  small    compared  to  r,   equal   to 

/  02  d  a,  which  is  the  moment  of  inertia  of  the  cross-section  I.     Also, 

da 

7^  = 


r 
18 


274:  SUPPLEMENT   TO    CHAP.  XIV.  [CHAP.  I. 

or   T  I  — —  =    I  da—  -  I  vda  +  -  I  =  A  -f  —3,  because   since 

1     J  r  +  v      J  rj  rjr  +  v  r* 

v  is  measured  from  the  centre  of  gravity,  /  v  d  a  =  0. 

rvda        f*  fvzda  I 

Again,  r  I  — —  =    I  v  a  a  —    I  — • —  =  —  — -. 

/    T  +  1)  I  I      T  +  V  T 

Therefore  the  insertion  of  these  values  of  the  integrals  in  the  equations 

for  _  and  _  above  gives 
E          E 

Gr       Ads  r~Ad(f>      AflJsnl 

E  ~"    ds         ~  [_  d  s     ~  r  ds\  r 

M        rA  d  ^      A  d  «~j  I 
E  r~  L  d  s         rdsj  r' 


M 


G-       A  a  s 
Inserting  the  second  in  the  first  of  the  above  equations,  ^-  =  — v-  A  —  =— , 

and  hence 


_-  — -  +  — —-  .     (8) 

ds        EA  +  EAr 

Inserting  this  in  the  second  equation  above, 


M  M  G 

E  I  +  E~A~^  +  E~A~7 


(c)  Change  of  length  and  position  of  axis. 

From  (8)  we  have  at  once  for  the  elongation  of  axis, 


or,  when  v  is  very  small  compared  to  r, 

1 


From  (9)  we  have  for  the  change  of  direction  of  the  tangent  to  the  axis 

A^  =  E/>(T  +  A~7*  +  A^r)d8'  or  from  (10)  for  r 
constant,  that  is,  for  a  circle, 


When  D  is  very  small  compared  to  r,  we  have 

ds    .......     (13) 


CHAP.  I.]  SUPPLEMENT   TO    CHAP.  XIV.  275 

If  the  piece  had  been  originally  straight,  d  A  (j£>  would  be  equal  to  d  <£,  and 

dA(b       dd>       1  El 

—3-^  =  —  —  =  -;  hence  from  (13)  we  have  M  =  -  . 

as        rd         r  r 


From  the  calculus  we  have  the  radius  of  curvature 

r     --  *       '  or>  aPProximatelv>  r  = 


hence  M  =  B  I  ~    .     .     .     .  -  .     .     .     (13  &) 

This  is  the  equation  assumed  in  the  Supplement  to  Chap.  XIII.  ,  Art.  1. 
2.  Displacement  of  any  point.  —  We  indicate  the  horizontal  dis- 
placement of  any  point  of  the  axis  along  the  axis  of  a  by  A  x,  along  y  by 
A  y.     The  corresponding  changes  of  d  x,  d  y,  and  d  s  are  A  d  x,  Ady,  Ads. 
The  total  horizontal  displacement  is  then  dx  +  Adx  =  (ds  +  Ads)  cos 
A  <£).     The  total  vertical  displacement  is  d  y  +  A  d  y  =  (d  s  +  A  d  s)  sin 
+  A  $).     Hence,  since  Adx  =  d  A  z, 


d  Ax  —  (d  s  +  Ads)  cos  (<£  +  A$)  —  dx, 
n  (0  +  A  </>)  —  d  y. 


By  Trigonometry,    cos  (0  +  A  $)  =  cos  </>  cos  A  <£  —  sin  <£  sin  A  0, 
sin   (0  -f  A  0)  =  sin  0  cos  A  <£  +  cos    0   sin   A  <£,   or  if    cos   A  <£  =  1, 

<?«  ^  <w 

sin  A  0  =  A  <£,  cos  </>  =  -r-  ,     sin  <£  =  -=—  . 


Substituting  these  in  the  equations  above, 

.  /      Ads\ 

dAx  =  (dx  —  A(pdy)  II  +—5 —  I  —  d  x, 
\  I 

(Ads\ 
1  +  — j —  I  —  d  y, 
as  / 

or,  removing  the  parentheses,  and  neglecting  quantities  of  the  second  order 
with  respect  to  A  (f>  and  — = — , 

dAx=  —  A<f>dtM — ^ —  dx 


-di~dy 


276  SUPPLEMENT   TO   CHAP.  XIV.  [CHAP.  T. 

When  the  radius  of  curvature  is  very  great  with  reference  to  the  thick- 
ness of  the  beam,  and  the  relative  change  of  length  — , —  is  disregarded, 

CL  8 

we  have  simply 


dAy  =  A<}>dx. 
But  from  (12)  A  <£  is  equal  to   /    _  _ 


Mds      Ads 

H 


/WLds 
E  j  ,  hence,   for  »  very  small  with 

reference  to  r, 


We  shall  have  frequent  occasion  to  refer  to  formulae  (8),  (12),  (13),  (14) 
and  (15)  in  the  following  discussion. 


CHAP.  H.]  SUPPLEMENT   TO   CHAP.  XTV.  277 


CHAPTER    II. 

HINGED  ARCH  IN   GENERAL. 

3.  Notation  —  Tlie  outer  fbrce§  in  general.  —  We  suppose  the 
ends  of  the  arch  to  be  hinged  at  the  abutments  at  the  centre  of  gravity  of 
the  end  cross-sections.  Then  the  end  reactions  must  pass  through  these 
points.  These  end  reactions  and  the  loads  constitute,  then,  the  outer  forces. 
For  equilibrium,  then,  the  horizontal  components  of  these  reactions  must 
be  equal.  Each  of  these  components  we  call  the  horizontal  thrust. 

We  use  the  following  notation  [PL  23,  Fig.  91]  : 

R  and  R',  the  reactions  at  ends  A  and  B. 

V  and  V,  their  vertical  components. 

H,  their  horizontal  component,  or  the  thrust. 

<z,  the  half  span. 

A,  the  rise  of  arch. 

a,  the  half  central  angle,  if  arch  is  circular. 

The  origin  of  co-ordinates  we  take  at  crown,  x  horizontal,  y  vertical.  The 
angle  of  radius  of  curvature  at  any  point  with  y,  or  of  tangent  to  curve  at 
any  point  with  #,  we  call  0. 

THE   OUTER   FORCES  IN  GENERAL. 

Suppose  a  single  load  P  to  act  at  any  point  E.     Let  its  horizontal  dis- 
tance from  crown  be  2,  the  corresponding  central  angle  E  O  C  be  /3. 
Then  the  conditions  of  equilibrium  are  : 

V  +  V  '  =  P, 

Va  —  V  a  —  Pz  =  Q. 

From  these  last,  we  have 


(16) 


For  a  circular  arc,  since  a  =  r  sin  a,  z  =  r  sin  0, 


sma  +  sinff  sma-sin/3 

2sina       '   I  ~~ 


We  distinguish  three  segments  in  the  arch  [Fig.  91],  viz.,  A  E,  or  from  end 
to  the  load  ;  E  O,  or  from  load  to  crown  ;  C  B,  or  from  crown  to  right 
end.  Quantities  referring  to  the  second  we  indicate  by  primes,  to  the 
third  by  double  primes.  Thus  for  the  tangential  component  of  the  result- 
ant at  any  point  within  A  E,  we  put  G  ;  for  the  normal  component,  N. 
In  E  O,  then,  we  have  G'  and  N'  ;  in  O  B,  G"  and  N*.  We  have  then 


278  SUPPLEMENT   TO   CHAP.  XIV.  [CHAP.  II. 

G  =  —  H  cos  (f>  —  V  sin  <£,  G'  =  —  H  cos  0  +  V  sin  $  )          ^ 
N  =  —  H  sin  0  +  V  cos  $,  N'  =  —  H  sin  0  —  V  cos  $  ) 

M  =  H(fc-y)-V(0-aO,  M'  =  H(fc-y)-  V  («  +  *)  •  •  (19) 
In  the  case  of  a  circular  arc,  a  =  r  sin  a,  h  =  r  (1  —  cos  a),  #  =  /•  sin  <£, 
y  =  r  (1  —  cos  (/>),  and  hence 

M  = 
JVT 


=  H  r  (cos  0  —  cos  a)  —  V  r  (sin  a  —  sin  <£)  ) 
=  H  r  (cos  0  —  cos  a)  —  W  (sin  a  +  sin  <£)  j 


4.  Intersection  Line.—  We  call  the  locus  of  d  [PL  23,  Fig.  91],  or 
the  curve  cdeik,  the  intersection  line. 

If  now  there  are  three  hinges,  one  at  crown  and  one  at  each  abutment, 
then  the  resultant  for  each  half  must  pass  through  the  crown  O.  If,  there- 

fore, for  the  crown  (*=0,  y=0),  we  make  in  (19)  M'=0,  we  have  H=V  -, 
or  inserting  the  value  of  V  from  (16), 

H  =  ^|^   .........    (21) 

If  the  load  lies  to  the  left  of  O,  then  only  the  resultant  If  acts  upon  the 
right  half,  and  must  pass,  as  above,  through  the  crown  O.  The  point  d 
lies  then  always  upon  B  C  or  A  O  prolonged.  Hence,  the  intersection  lines 
are  two  straight  lines,  which  pass  through  the  crown  and  ends. 

5.  Parabolic  Arc—  concentrated  Load.  —  For  a  parabola  we 

have  y  —  —  xz  ;  hence,  d  y  —  —  x  d  x,  and,  approximately,  ds  =d  x. 
a'  a 

(a)   Change  of  direction  of  tangents. 

Inserting  this  value  of  y  in  equations  (18)  for  M  and  M',  we  have  from 
equation  (13),  Art.  1,  since  r  d  0  =  d  s  =  d  x  for  the  change  of  direction  of 
the  tangent  at  any  point  before  and  after  flexure, 


Integrating  this,  we  have  for  the  three  segments  A  E,  E  C  and  O  B, 


.  .    (22) 


where  A,  A',  A"  are  constants  of  integration,  to  be  determined  by  the  as- 
signment of  the  proper  limits.  Thus,  if  we  make  x  =  z,  the  two  first  of 
equations  (22)  are  equal;  hence, 


CHAP.  II.]  SUPPLEMENT   TO    CHAP.  XIV.  279 

and  accordingly  A  -  A'  =  (V  -  V)  a  z  -  \  (V  +  V)  z.     Since  V  +  V  =  P, 
and  from  (16)  V  —  V  =  P  —  ,  we  have, 

a 

A-A'=lPz*     .......     (I.) 

(&)  Horizontal  displacement. 

Inserting  the  value  of  dy  for  the  parabola,  viz.,  ay  =  —  xdx,  and  the 

value  of  M  from   (19),  and  inserting  in  this  last  the  value  of  y,  viz., 

y  =  —  x*,  we  have  from  equation  (15),  Art.  2, 
2 


Integrating  this,  we  have  for  the  three  divisions  of  the  arch,  as  before, 


For  the  point  B,  or  x  =  z,  we  have  from  the  two  first  of  these  equations, 
B  -  B'  =  1  (V  -  V)  a  z*  -  i  (V  +  V)  z4  -  £  (A  -  A')  z*,  that  is, 

B-B=-2\-Ps*    .......     (II.) 

For  the  crown,  x  =  0,  and  the  second  and  third  equations  are  equal, 
hence 

B'  =  B*    ........     (III.) 

For  the  left  end,  that  is,  for  x  =  a,  since  the  end  of  the  arch  must  not  slip, 
we  must  have  x  —  x  =  0.  So  also  for  the  right  end,  for  x  =  —  a.  There- 
fore, from  the  first  and  third  equations,  putting  B'  for  B",  we  have 

&Ua3h--*sVa4'  +  iAa2  +  B  =  0, 
-  A-  H  a3  h  +  -fr  V  a4  +  i  A"  a2  +  B'  =  0. 

By  the  addition  and  subtraction  of  these  equations,  we  have,  since 
V  +  V  =  P,  and  (V  -  V')  a  =  Pz, 

A")»2-(B  +  B')  =  0    ....     (IV.) 
^  +  24)  +  i(A-A")a2=0    .    .     (V.) 

We  might,  in  a  precisely  similar  manner,  form  three  equations  similar  to 
(23)  for  the  vertical  displacement  A  y.  This  would  introduce  three  more 
constants  and  four  more  equations  of  conditioh  between  them.  By  the 
nine  equations  I.  to  IX.  thus  obtained,  these  constants  may  be  then  deter- 
mined in  terms  of  the  known  quantities  H,  A,  P,  a  and  z,  and  thus  the 
change  of  shape  at  any  point  may  be  fully  determined. 

The  complete  discussion,  as  indicated,  is  unnecessary  for  the  purpose  we 


2SO  SUPPLEMENT   TO    CHAP.  XIV.  [OHA.P.  II. 

have  in  view,  and  we  shall  not,  therefore,  pursue  it  further.  We  have 
already  all  the  general  formulae  of  which  we  shall  need  to  make  use  in  the 
discussion  of  the  parabolic  arch. 

6.  Circular  Arc  —  concentrated  Load.  —  In  a  perfectly  similar 
manner  we  may  make  out  analogous  formulae  for  the  circular  arch.  Thus, 
referring  to  equation  (8),  Art.  1,  and  inserting  for  G  and  M  their  values  as 
given  in  (18)  and  (19),  Art.  3,  we  have  for  the  force  in  the  direction  of 
the  axis  (see  eq.  4), 


. 
E  A  -  =  —  H  cos  a  —  V  sin  a 


<E  A  -  —  =  —  H  cos  a  —  V  sin  a 

ds 

Putting  for  Hr  Vand  V  their  values  from  (21),  Art.  4,  and  (17),  Art.  3,  we 
have, 

A  d  s  _      _  sin  a  +  sin  /3  —  2  cos  a  sin  $ 

= 


.    .    (25) 
A  d  s'  _       —  sin  a  —  sin  /3 

~dT'  2(1  -COS  a) 

(a)  Change  of  direction  of  tangents. 

Referring  to  equation  (12),  Art.  1,  we  have,  since  rd<f)  =  ds  and  r  </>  =  «, 

/M  Ads 

EIr**  +  -dT*- 

The  value  of  M  is  given  in  (20),  of  ^U-J  in  (24).    Inserting  these  values, 
we  have 

A  <£  =  ^j-j  /  |~H  (cos  $  —  cos  a)  —V  (sin  a  —  sin 

—  ==-r-  (H  cos  a  -H  V  sin  a)  $. 

Performing  the  integration,*  and  putting,  for  brevity,  *  =  —  —  -,  we  have  for 
the  three  segments  of  the  arc,  as  before, 

ElA£  =  r2  pK  (sin  <£  -  <£  cos  a)  -  V  Casino  +  cos  <f>)~^  -  K  T*  (H  cos  a  +  V  sin  a)  $  +  A  "| 
El  A<£'  =ra  [~H  (sin</>—  £cosa)  —  V(<£sina—  cos  </»)"!  —  K  r2  (H  cos  a  +  Vsina)0  +  A'   t  26 
E  I  A  $"  =  r*  [H  (sin  <£-  0  cos  a)  -  V  (<£  sin  a  -  cos<£)]-  K  r2(Hcos  a  +  V  sin  a)<f»  +  A"  J 

For  the  point  E,  <£  =  3,  and  the  first  and  second  equations  become  simul- 
taneous.    Hence,  after  reduction, 

A-A'  =  (V-  V)  r2/3sina  +  (V  +  V)  r*  cos  0  +  K  (V-V)  rz  /3sin«. 


But  V  +  V  =  P,  and  from  (6)  V  -  V  =  P         ,  hence 

sin  a 


.     .     .     (I.) 


.  *  See  Art.  7,  f  allowing. 


CHAP.  II.]  SUPPLEMENT   TO   CHAP.  XIV. 

(b)  Horizontal  displacement. 

According  to   eq.    (14),  Art.    2,   since    x  =  r  sin  0,   y  —  r  (1  —  cos  0), 
dx  =  r  cos  0  d  0,  d  y  =  r  sin  0  d  0,  we  have 

E  I A  x  =  r    I  [H  r'2  (sin  0  —  0  cos  a)  —  W  (0  sin  a  +  cos  0), 

—  K  r*  (H  cos  a  +  V  sin  a)  0  +  A  J  sin  0  <Z  0, 

—  K  r2    /  (H  cos  a  +  V  sin  a)  cos  0  d  0. 
The  integration  gives  us  the  three  equations, 

E I A  x  =  —  r*    H  ( £  0  —  \  sin  0  cos  $  —  cos  a  sin  0  +  0  cos  a  cos  0) 

—  V  (sin  a  sin  0  —  0  sin  a  cos  0  +  •£  sin2  0)1 

—  K  r3  (H  cos  a  +  V  sin  a)  0  cos  0  —  A  r  (1  —  cos  0)  +  B. 

£  I A  #'  —  —  r3 1  H  (•£  0  —  -J  sin  0  cos  0  —  cos  a  sin  04-0  cos  a  cos  0) 

—  V  (sin  a  sin  <^>  —  0  sin  a  cos  (f>  —  i  sin2  0)  I 

—  K  r3  (H  oosa  +  V  sin  a)  0  cos  0  —  A'  r  (1  —  cos  0)  +  B'. 

E I A  x"  =  —  r*  I  H  (f  0  —  |  sin  0  cos  0  —  cos  a  sin  0  +  <j>  cos  a  cos  0) 

—V  (sin  a  sin  0  —  0  sin  a  cos  0  —  \  sin2  <^>) 

—  K  r3  (H  cos  a  +  V  sin  a)  0  cos  0  —  A"  r  (1  —  cos  0)  +  B". 

For  0  =  £,  that  is  at  the  load,  A  a?  must  equal  A  x'. 
Hence,  after  reduction, 

B  -  B'  =  -$Pr3  (2  +  sina/3-  2003/3-2)8 sin /3)  +  * P r3 /3 sin /3  .  .  (II.) 
For  the  crown  0  =  0,  and  A  x'  —  A  x" ;  hence 

B'  =  B" (III.) 

For  the  left  end,   0=a  and  A»  =  0;    for  the  right  end,  0  =  —  a  and 
A  a"  =  0;  that  is, 

—  iHr3  (a  —  3  sin  a  COS  a  -f  2aCOS2a)  4- i  V  r3  (3  sin2  a  —  2  a  sin  a  COS  a) 
—  K  r3  (H  cos  a  +  V  sin  a)  a  cos  a  —  A  r  (1  —  cos  a)  +  B  =  0,  and 

4-  i  H  r3  (a  —  3  sin  a  cos  a  +  2  a  cos2  a)  —  |  V  r3  (3  sin2  a  —  2  a  sin  a  cos  a) 
•f  K  y3  (H  cos  a  +  V  sin  a)  a  cos  a  —  A"  r  (1  —  cos  a)  +  B"  =  0. 

The  subtraction  and  addition  of  these  equation  gives,  after  reduction, 

H  r2  (a  —  3  sin  a  cos  a  +  2  a  cos2  a) 
—  iPr2  (3  sin2  a  —  2  a  sin  a  cos  a  —  2  —  sin2  3  +  2cos/3+  2/3sin/3) 

+  K  rz  12  H  a  cos2  a  +  P  (a  sin  a  cos  a  —  /3  sin  /3)  I 

+  (A-A")(l-cosa)  =  0 (IV.) 

and  -J  Pr3  sin  3  (3  sin  a  —  2  a  cos  a) 

-  (A  +  A")  r  (1  -  cos  a)  -  K  Pr3a  cosa  sin/3  +  B  +  B'  =  0  .   .  (V.) 


282  SUPPLEMENT   TO    CHAP.  XIV.  [CHAP.  II. 

Here,  as,  before,  we  shall  leave  the  discussion,  as  we  have  already  all  the 
equations  of  which  we  shall  make  use. 

7.  Integrals  used  in  the  above  Discussion.— For  convenience 
of  reference,  we  here  group  together  the  known  integrals  employed  in  the 
preceding  discussion. 

/  sin  x  d  x  =  —  cos  x,       /  cos  x  d  x  —  sin  x. 
I  sin2  ado:  =  $x  —  isinxcosx,      I  cos2  xdx  =  ix  +  •£  sin  x  cos  x. 

I  sin  x  cos  x  dx  —  i  sin2  x. 

I  sin3  x  dx  —  —  \  cos  x  (2  +  sin2  x),      I  cos3  x  d  x  =  $  sin  x  (2  -+-  cos2  x). 
I  sin2  x  cos  x  dx  =  $  sin3  x,      I  sin  x  cos2  x  d  x  =  —  $  cos3  x. 
I  xsmxdx  =  smx  —  x  cos  x,    I  x  cos  x  d  x  =  cos  x  +  x  sin  x. 
I  x  sin2  xdx  =  ±x*+i  sin2  x  —  |  x  sin  x  cos  x. 
I  x  cos2  xdx  =  ±x*  —  i  sin2  x  +  %x  sin  #  cos  «. 
I  xsinx  cos  #  tZ  a?  =  J  (2  «  sin2  x  —  #  +  sin  x  cos  #). 


CHAP.  III.]  SUPPLEMENT  TO   CHAP.  XIV.  283 


CHAPTER  III. 

AEOH   HINGED   AT   ABUTMENTS    ONLY  -  CONTINUOUS   AT   CROWN. 
A.    PARABOLIC  ARC  —  CONSTANT   CROSS-SECTION  —  CONCENTRATED  LOAD. 

8.  Horizontal  Thru§t.—  We  can  apply  here  directly  the  results  of 
Art  5.     Thus,  in  equations  (22)  for  x  =  0,  A  $  =  0,  and  A  0"  =  0,  hence 
A'  =  A".     If  then  in  eq.  (V.)  of  that  Art.  we  put  A'  for  A",  and  then  for 
A  —  A'  its  value  from  (I.),  we  have  at  once 

H_  5  p5a*-6aV  +  g*_  (5  a*  -  z*)  (a*  -  z>) 

H=r*P~          a*  h  ~64P  ~^h~  "(27) 

This  is  the  formula  which  we  have  given  in  Art.  159  of  the  text,  without 
demonstration.     The  thrust  is  greatest  when  the  load  is  at  the  crown.     We 

have  then  z  —  0  and  H  =  |f  P  %    The  value  of  V  is  given  in  Art.  3. 

n 

9.  Inter§ection  Curve.  —  Denote  the  ordinate  of  the  curve  cdei/k 
(PL  23,  Fig.  91),  taken  above  the  line  A  B  by   y.     Then  we  see  from  the 

•y 

Fig.  that  y  =  A  N  tang,   d  A  N  —  (a  —  z)  =.     The  value  of  H  is  given 

above,   that  of   V  has    already  been   found  in  Art.  3,   eq.  (16).  viz., 
V  =  P  -  .    Hence  we  have,  after  reduction, 


™ 
(28) 


which  is  the  equation  already  given  in  Art.  159,  and  from  (18)  and  (19) 
the  values  of  the  table  in  that  Art.  have  been  calculated. 

The  above  values  of  H  and  V  are  simple  and  of  easy  application,  not  in- 
volving much  calculation  in  any  special  case.  Hence  we  can  readily  com- 
pute H  and  V,  and  thus  check  the  accuracy  of  our  method  of  construction 
given  in  Chap.  XIV. 

B.    CIRCULAR  ARC  —  CONSTANT    CROSS-SECTION  —  CONCENTRATED  LOAD. 

1O.  Horizontal  Thru§t.  —  Here  we  can  apply  directly  the  results  of 
Art.  6.  Thus,  inserting  in  eq.  (IV.)  of  that  Art.  A  —  A'  for  A  —  A",  and 
taking  the  value  of  A  —  A'  from  (I.),  we  have  an  equation  for  the  deter- 
mination of  H.  This,  after  reduction,  becomes 

_      sln2  a—  sin-  ft  +  2  cos  a  (cos  /3  —  cos  a)  —  2  (1  +  K)  cos  a  (a  sin  a  —  )3  sin  g) 
2  [a  —  3  sin  a  cos  a  +  2  (1  +  *)  a  cos2  a] 

which  is  the  equation  given  in  Art.  159  (2)  of  the  text. 

For  the  semi-circle,  a  =  90°,  sin  a  —  1,  and  cos  a  =  0,  and  this  becomes 


284:  SUPPLEMENT   TO    CHAP.  XIV.  [CHAP.  III. 

If  we  put 

A!  =  sin2  a  —  sin2  £  +  2  cos  a  (cos  /3  —  cos  a  —  a  sin  a  +  /3  sin  j3), 
A2  =  2  cos  a  (a  sin  a  —  /3  sin  $), 
B!  —  2  (a  —  3  sin  a  COS  a  +  2  a  COS2  a), 
B2  =  2  a  COS2  a, 
we  have 

A1-A2^_      *~A/  /A, 
Bx+B,    •    *V       B,      \B, 

H* 

A  TJ  A 

or,  if  we  put  A  =  -^,  B=  g2,  and  H0  =  P  ~,  we  have 

,  1  —  A  K 


But  Ho  =  P  .g-1  is  the  value  of  H  from  the  formula  above,  when  the  terms 

containing  Ic  are  disregarded. 

We  have  also,  by  series  (see  Art.  20,  following), 

A!  =  -,V  («2  —  £2)  [  (5  a2  -  /32)  -  ^o  (49  a4  +  34  .a2  /32  -  11  £*)  +  .  .  .] 
A2  =  2  (a2  -  /32)  [1  -  i  (4  a2  +  ^2)  +  .   .   .] 
B:  =  ^  a6  (1  -  -/r  a2  +  .   .   .)          B2  =  2  a  (1  -  a2  +  .   .    .) 
Approximately,  therefore,  since  for  h  small  with  respect  to  a,  the  tan- 
gent may  be  taken  for  the  arc,  and  hence  —  =  — ,  or  a  =  — ,  we  have 

rah  2  h 

24  6  a2 15  _  15  a* 

~~- 


Hence,  when  rise  is  small  compared  with  span,  we  have  the  approximate 
expression 


.    24     - 


5  A2     1--  - 


15  15 


By  means  of  a  table  calculated  for  H0,  for  various  values  of  a  and  /3  =  0, 
0.2,  0.4,  etc.,  of  a,  the  thrust  can  be  readily  found  in  any  case  from  the 
above  formula. 

We  give  in  the  following  Tables  the  values  of  H, ,  A  and  B,  calculated 
from  the  exact  formulae.  The  formula  for  H  above  is  thus  made  of  easy 
practical  application,  without  tedious  calculation,  and  the  results  given  by 
the  method  of  Chap.  XIV.  may  easily  be  checked. 

The  value  of  V  is  given  in  Art.  3. 


CHAP.  HI.] 


SUPPLEMENT   TO    CHAP.  XIV. 

TABLE  FOB  H0. 


285 


a 

a  =  0 

a  =  10° 

a  =  20° 

o  =  30° 

a  =  40° 

a  =  50° 

a  =  60° 

a  =  90° 

0 

0.391 

0.391 

0.388 

0.385 

0.380 

0.373 

0.364 

0.318 

0.2 

0.372 

0.372 

0.369 

0.365       0.359 

0.352 

0.342 

0.288 

0.4 

0.318 

0.317 

0.315 

0.309 

0.301 

0.292 

0.278 

0.208 

0.6 

0.232 

0:231 

0.228 

0.222 

0.213 

0.202 

0.187 

0.110 

0.8 

0.123 

0.122 

0.119 

0.115       0.108 

0.099 

0.086 

0.030 

1 

0 

0 

0 

0 

0 

0 

0 

0 

a 

Coefficients  of     P^. 
h 

Thus,  for  /3  =  0, 


,  etc.,  of  a,  the  numbers  in  the  table  give  the  co- 


efficients of  P  —  for  a  =  0,  10°,  20°,  etc. 
h 

For  the  values  of  A  and  B,  we  have  the  following 
TABLE  FOR  A  AND  B. 


a 

gj 

a  =  10° 

a  =20" 

a  =30° 

a  =  40° 

a  =50° 

a  =  60° 

a  =90° 

i 

0 

1.20 

1.19 

1.17 

1.14 

1.08 

1.00 

0.88 

0 

Coeffi- 

0.2 

1.21 

1.20 

1.18 

1.15 

1.10 

1.01 

0.90 

0 

of 

A 

0.4 

1.24 

1.24 

1.21 

1.18 

1.13 

1.05 

0.94 

0 

a* 

0.6 

1.29 

1.29 

1.27 

1.24 

1.20 

4.13 

1.02 

0 

0.8 

1.38 

1.38 

1.36 

1.34 

1.30 

1.24 

1.18 

0 

1' 

1.50 

1.50 

1.49 

1.47 

1.45 

1.41 

1.36 

0 

B 

0.234 

0.233 

0.221 

0.203 

0.178 

0.146 

0.107 

0 

a4 

,0  2 

Thus,  for  various  values  of  — ,  we  have  the  coefficients  of  ^,  which  give 

A  for  a  =  10°,  20°,  etc.,  and  for  the  values  of  a  have  the  coefficients  of 

a4 

— ,  which  give  B. 

11.  Intersection  Curve.— Indicating,  as  before,  by  y  the  ordinate 


286  SUPPLEMENT    TO    CHAP.  XIV.                         [CHAP.  III. 

Nd  of  the  curve  cdeilc  [Fig.  91],   we  have,  as  before,   y=ANtan^. 

d  A  N  =  (a  -  z)  ^,  or  since  a  =  r  sin  a,   z  =  r  sin  fl.  V  =  P  sm  «  +  sin  ff 

1=1  2  sin  a 

from  eq.  (17), 


2  sin  a        H 
Inserting  the  value  of  H  above,  we  have 

sin2  a  —  sin2  /3  ~1  +  B  »     B  4 


2  sin  a  l  -  A  AC     A 

/sin2  a  —  sin'/SXBi 
or  if  yo  =  r  I  --  g-jjT^  -  1^-  ;  that  is,  if  y0  is  the  value  of  y  when  &  i 

neglected, 


which  is  the  value  of  y  given  in  Art.  159  (2)  of  the  text.  In  that  Art.  we 
have  already  tabulated  the  values  of  A  and  B,  as  also  of  y0  for  various 
values  of  a  and  /3. 

For  /3  =  a,  that  is,  for  the  end  ordinate,  our  expression  for  y  reduces  to 

—     In  this  case,  by  differentiating  numerator  and  denominator,  we  have 

a  —  3  sin  a  cos  a  +  2  (1  +  K)  a  cos"  a 
sin  a  —  a  COS  a  —  K  (sin  a  +  a  COS  a)' 

For  the  semi-circle,    a  =  90°  =  —  ,  sin  a  =  1,     cos  a  =  0,      and  hence 

2 

y  =  \  nr  =  1.5708  r.  Hence,  for  the  semi-circle  the  intersection  curve  be- 
comes a  horizontal  straight  line  at  0.5708  r  above  the  crown.  In  all  cases 
for  small  central  angle  c,  K  may  be  disregarded. 

The  above  results  are  sufficient  to  enable  us  to  either  diagram  or  calcu- 
late the  strains  in  every  piece  for  any  given  position  of  load. 


CHAJP.  IV.]          SUPPLEMENT  TO  CHAP.  XIY.  287 


CHAPTER    IV. 

ARCH     FIXED     AT     ENDS. 

12.  Introduction. —  In  the  previous  case,  the  end  reactions  pass 
always  through  the  ends.     If,  however,  the  ends  are  "  walled  in,"  so  that 
the  end  cross-sections  remain  unchanged  in  position,  and  cannot  turn,  these 
reactions  pass  then  no  longer  through  the  centres  of  the  end  cross-sections. 
In  the  first  case,  the  moments  at  the  ends  are  zero ;  now,  however,  we  have 
end  moments  to  be  determined,  viz.,  Mj  and  M2,  left  and  right.     For  their 
determination  we  have  the  condition  that  the  tangents  to  the  curve  at  the 
ends  must  always  remain  invariable  in  direction,  or  for  the  ends,  A  ^>  =  0. 

In  the  arch  above  with  hinges  at  ends,  we  have  always  considered  a  por- 
tion lying  between  the  end  and  any  point.  In  the  present  case,  however, 
we  shall  consider  the  portion  between  the  crown  and  any  point.  Both 
methods  lead,  of  course,  to  the  same  results,  but  the  latter,  in  the  present 
case,  is  somewhat  simpler. 

Accordingly,  we  conceive  the  arch  cut  through  at  the  crown  [PI.  24, 
Fig.  93].  The  total  resultant  force  exerted  upon  the  one-half  by  the  other, 
we  decompose  into  a  vertical  force  V  at  the  crown,  and  a  horizontal  force 
H.  The  distance  c  &  of  this  last  from  the  centre  of  gravity  of  the  section 
at  crown  is  e0,  and  hence  the  moment  at  crown  is  M0  —  —  H  e0. 

13.  Concentrated  Load— General  Formulae. — Let  a  weight 
P  act  at  any  point;  then  representing,  as  before,  by  primes,  quantities 
relating  to  the  portion  between  the  load  and  right  end,  we  have,  as  in  (18), 

G  =  —  H  cos  <j>  —  (P  —  V)  sin  0,    G'  =  —  H  cos  0  +  V  sin  <£  | 

N ;  =  -  H  sin  0  +  (P  -  V)  cos  0,    N'  =  -  H  sin  $  -  V  sin  $  \"  ^ 

Also,     M  =  -  H  (e0  +  y)  -  Vx  +  P  (x  -  z),     M'  =  -  H  (e0  +  y}  -  V  x, 

or,  since     —  H  e0  —  M0  =  moment  at  crown, 

M  =  Mo  -  H  y  -  V  x  +  P(x  -  «),     M'  =  M0  -  H  y  -  V  x  .   .  (32) 

(a)  Intersection  curve. 

The  two  reactions,  R  and  R',  intersect,  as  before,  in  a  point  L  (Fig.  92), 
which  must  lie  upon  P  prolonged,  as  otherwise  R,  R'  and  P  could  not  be 
in  equilibrium.  The  locus  of  the  point  L  we  call,  as  before,  the  intersection 
curve.  The  equation  of  this  curve  can  be  easily  found  when  V,  H  and  M0 
are  known. 

The  force  acting  upon  the  portion  B  E  (Fig.  92)  is  the  resultant  of  V 
and  H.  The  component  H  acts  at  the  point  of  intersection  o  of  this 
resultant  L  ^  with  the  vertical  through  C.  The  vertical  distance  of  this 
point  o  from  c  is,  as  above,  e0 ;  its  horizontal  distance  from  P  is  z.  Then 
z  cot  L  ^  Tc  is  the  vertical  distance  of  this  point  from  L,  and  e0  4-  z  cot 


288 


SUPPLEMENT   TO   CHAP.  XIV. 


[CHAP.  iv. 


=  e0  +  z  ~  =  y,  where,  as  in  the  Fig.,  e0  is  negative.     In  any  case,  e0 

IB 

jyr 

is  given  by  e0  =  — ^F  5  hence,  for  the  intersection  curve, 


H 


(33) 


(&)  Direction  curves  and  segments. 

The  direction  of  the  resultants  R  and  R'  can  be  determined  in  two  ways 
First,  by  the  points  of  intersection  $  and  ^  with  the  verticals  through  the 
centres  of  the  end  cross-sections ;  second,  by  means  of  the  curves  enveloped 
by  these  resultants  for  every  position  of  P.  We  call  the  first  distances 
A  (f)  =  Ci,  B  \|r  =  c2,  the  direction  segments,  and  the  enveloped  curves  the 
direction  curves. 

Taking  Ci  and  c2  as  positive  when  laid  off  upwards  above  the  ends,  we 
have  MI  =  —  H  c\     M2  =  —  H  c2 ;     therefore 


(34) 


We  may  also  easily  determine  the  equation  of  the  direction  curves.  Let 
the  co-ordinates  with  reference  to  the  crown  of  any  point  be  v  and  w  (Fig. 
92).  If  the  load  P  is  now  moved  through  an  indefinitely  small  distance, 
the  new  resultant  cuts  the  former  in  a  point  of  the  curve  required.  These 
two  resultants  intersect  the  vertical  through  C  in  two  points.  Let  the  dis- 
tances of  these  points  from  O  be  c  and  c  +  d  c,  and  let  y  and  y  +  d  y  be 
the  angles  of  the  resultants  with  the  vertical.  Then  v  —  (w  +  c)  tan  y, 
v  =  (w  +  c  +  d  c)  tan  (y  +  d  y). 

Eliminating  v, 

(c  +dc)  tan  (y  +  d  y)  —  c  tan  y  _       d  (c  tan  y) 
tan  (y  +  d  y)  — •  tan  y  d  tan  y 

From  the  first  of  the  equations  above  we  have  then 

dc 

v  =  —  -= tan2  y. 

d  tany 


But  tan  y  =  -,   c  =  -  -,   c  tan  y  =  —  -,  hence 


.    .     (35) 


w  = 


d 


Thus  we  see  that  in  any  case  we  have  only  to  determine  H,  V  and  M0,  and 
we  can  then  from  (33)  and  (34)  or  (35)  determine  at  once  the  intersection 


CHAP.  IV.]          SUPPLEMENT  TO  CHAP.  XIV.  289 

curve,  and  the  direction  segments  or  curves.  These  are  all  we  need  for  our 
method  of  construction  as  given  in  Chap.  XIV. ;  once  given,  we  can  then 
easily  construct  H,  V  and  M9  or  Mi  for  any  position  of  weight. 

A.    PARABOLIC  ARC — CONSTANT   CROSS-SECTION — CONCENTRATED  LOAD. 

#2 

14.  Determination    of   H,    V    and  Mo.— We    put    y  =  h  — , 

dy  =  2  -jdx,  ds  =  dx,  as  before.    Then  from  the  values  of  M  given  in 

(32)  we  have,  according  to  equation  (13),  Art.  1,  after  inserting  the  values 
of  y  and  d  s  above,  and  integrating, 


and 

r-  T3T  7.  -i 

+  A'. 


For  x  =  z,   A  $  =  A  $',  hence  |  P  (z  —  2  z)  z  +  A  =  A',  or 

A-A'  =  iPs2     .......      (L) 

For  x  =  a,   A  $  —  0,   and  for  x  =  —  a,  A  0'  =  0,  hence 

0  =  a 


0  =  -ajMo  -iHA  +  *Vo     +  A', 

and  by  addition  and  subtraction, 

A  +  A'  =  Va't-lP(a-2z)a      ....      (II.) 

2M0a-fHaA  +  iP(a-2)2  =  0      .     .     .      (IE.) 
From  I.  and  II.  we  have 

A  =  |V«2-iP(a2  -  2as-22)  ) 

A'^-LVa2~iF(a2-2as  +  s2)  j 

For  the  horizontal  and  vertical  displacement  of  any  point,  we  have  from 
(  5),  Art.  2,  after  integration, 


E  I  A  x  =-          i  Mo  z3-          rf-i  Vz<+iP  Q^-  |  sz3)  +  i  A  «2  +  B 


], 
J 


El  Art  =-  ^-[i  Mo  z3-^  V-*  V  «4+i  AV+B'l 
and 

E  I  A  y  =  iM0  «2-  ?^2  ^4-|  (V-P)  z'- 
i«  a 

B  L  A  y»=  i  Mo  x9-  ?s  a;4-i  V  aj«+  A' 
19 


290 


SUPPLEMENT   TO    CHAP.  XIY. 


[CHAP.  rv. 

For  x  =  z,   A  x  =  A  a?',   and  A  y  =  A  y',   hence 

B-B'  =  /4-Ps4-i(A-A>2,     and   C  -  C' =  $P  zs  -  (A  -  A')z,    or 

(IV.) 
(V.) 

For    a  =  a,   A  a?  =  0,   A  y  =  0,    and    f or    35  =  —  a,     A  *'  =  0,     A  y'  =  0. 
hence, 


0  =  +  I  Mo  a2  -  nfeH  Ji  a*  -  i  (V  -  F)a3  -  $  Pzo?  +  Aa  +  O, 

0  =  +  i  Mo  a2  —  ff  H  h  a2  +  %  V  a3  —  A'  a  +  O'. 

The  addition  and  subtraction  of  the  two  first  and  two  last  of  these  equa- 
tions gives,  when  we  put  for  A  +A',  A  —  A',  B  —  B',  O  —  C',  their 
values  above : 

2  M0  a3  —  |  H  Ji  a3  +  •£  P  (3  a4  —  8  a3  z  4-  6  a2  s2  —  24)  =  0  .  .  (VII.) 

O  +  C'  =  —  Mo  a?  +  |  H  h  a?  —  £  P  a  (a8  —  3  a  z  +  3  s2)  . .  (VIII.) 

4  f  V  a9  —  i  P  (2  a9  —  3  a8  s  +  z*)  =  0 (IX) 

Equations  HE.  and  VII.  contain  only  H  and  M0  unknown.     Their  solu- 
tion gives 

(36) 


a3  h 


M0  =  - 

From  IX.  we  find  directly 


.    .     .     (37) 


(38) 


These  are  the  equations  given  in  Art.  160.     From  them  we  have  the  fol- 
lowing table : 


z 

H 

V 

Mo 

z 

H 

V 

Mo 

0 

0.4688 

0.5000 

-  0.09375 

0.5 

0.2637 

0.1563 

+  0.02539 

0.1 

0.4594 

0.4252 

-  0.04936 

0.6 

0.1920 

0.1040 

+  0.02400 

0.2 

0.4320 

0.3520 

-  0.01606 

0.7 

0.1219 

0.0607 

+  0.01814 

0,3 

0.3882 

0.2818 

+  0.00689 

0.8 

0.0607 

0.0280 

+  0.01025 

0.4 

0.3308 

0.2160 

+  0.02025 

0.9 

0.0169 

0.0073 

+  0.00314 

0.5 

0.2637 

0.1562 

+  0.02539 

1.0 

0. 

0. 

0. 

.a 

•*f 

.P 

.Pa 

.a 

•-: 

.P 

.Pa 

CHAP.  IV.]  SUPPLEMENT   TO    CHAP.  XIV.  291 

From  (34)  and  (35)  we  can  now  find  the  intersection  and  direction  curves. 
The  preceding  table  gives  us  sufficient  data  for  complete  calculation  by 
moments  according  to  Art.  162.  The  intersection  and  direction  curves 
will,  as  already  explained,  enable  us  to  find  the  above  quantities  graphi- 
cally. 

15.  Intersection  Curve.  —  From  (33)  we  have  y  =  —         —  ,  or  m- 

H 

serting  the  values  of  H,  V  and  M0  above,  and  reducing, 

Hy  =  &P  (a  ~  ^  (a  +  z?  =  iHft,   hence  y  =  *fc 
a 

For  the  parabolic  arch  with  fixed  ends,  then,  the  intersection  curve  becomes 
a  straight  horizontal  line,  i  h  above  the  crown. 

16.  Direction  Curve.—  From  (36),  (37)  and  (38)  we  have 

-22g     dV          SP'-s8 


dz  ~  8a*h        '     dz  4  a3 

P  (a  -  z)  (4aa  -  5az  -  5g2) 


dz  Sa* 

Inserting  these  in  (35),  as  also  the  values  of  H,  V  and  M0  themselves,  and 
reducing,  we  have 


(39) 

15  (a  +  z)  (3  a  +  2) 

For  z  =  0,   0  =  |  a,  w  =  £f  h.    For  z  =  a,   v  =  %a,   w  —  £  h. 
For  z  —  —  a,   v  =  a,   w  =  —  oo .     Eliminating  s,  we  have 


x 
15  a  (a  —  v) 

This  is  the  equation  of  an  hyperbola.  Hence,  for  the  parabolic  arc  with 
fixed  ends,  the  direction  curve  is  upon  each  side  of  the  crown  an  hyperbola. 
This  hyperbola  is  described  in  Art.  160  of  the  text,  (Fig.  93),  and  a  table 
to  facilitate  its  construction  is  there  given. 

B.   CIRCULAR  ARC  —  CONSTANT    CROSS-SECTION  —  CONCENTRATED  LOAD. 

17.  Fundamental  Equations.  —  From  eq.    (32)   we  have,   since 

x  =  r  sin  $,   y  =  r  (1  —  cos  <£),   z  =  r  sin  0, 

M  =  Mo  —  Hr(l  —  cos$)  +  (P  —  V)rsin$  —  Ffsin/3 


'  =  M0  -  Hr(l  -  ' 


The  expressions  for  G,  Art.  13,  eq.  (12),  apply  here  directly. 
Therefore,  from  eq.  (8),  Art.  1,  we  have 


292  SUPPLEMENT   TO    CHAP.  XIV.  [CHAP.  IV. 

Hence  from  (12),  Art.  1,  since  ds  =  rdfa 

^rM0--Hr(l-cos0)  +  (P-V)rsin(/)-P7'sm/3]^0-f  —  d<p, 

-^rMo-Hr(l-cos<£)-Vrsind>l  dd)  +  Sds'  0. 
E  I L  J  d  s 

Substituting  the  values  of   above,  integrating,  and  putting,  as  be- 
fore, for  brevity,   K  =  —^  we  have 

E I A  <£  =  r  [MO  0  —  H  r  (<f>  —  sin  <£)  —  (P  —  V)  r  cos  $  —  P  r  $>  sin  /3  J 

+  K  r  (M0  -  H>  -  Gr  sin  /3)  $  +  A. 

E I A  <£' '  =  r  [MO  <j>-  Hr  ($  —  sin  <£)  +  Vr  cos  $J  +  K  r  (M0  —  H>)  $+A'. 
For  0  =  0,   A  ^  =  A  0',   and  we  obtain 

A  -  A'  -  Pr2  [cos  ft  +  (1  +  K)  j3  sin  £  J     .     .     .     (I.) 

For  <p  =  a,   A  <£  =  0,  and  for  <J>  =  —  a,  A  #'  =  0.     Adding  and  subtracting 
the  equations  thus  obtained,  and  eliminating  A  —  A',  we  have 

A  +  A'  =  P  r2 1  cos  a  +  (1  +  K)  a  sin  /3~|  -  2  V  r8  cos  a  .  .  (II.) 
2  Mo  a  -  2Hr(a-  sina)-Pr  [cos  a  -  cos  0  4-  (a  -  0)  sin  /3  J 

+  K  [2  Mo  a  —  2H  r  a  -  P  r  (a  —  j3)  sin  /3j  =  0    .  .  (HI.) 

From  eq.  (14),  Art.  2,  we  have,  as  before,  after  integrating,  for  the  hori- 
zontal displacement, 

E I A  a?— —Mo  ?*2  (sin  <j>— <p  cos  4>)+£  H>3  (2  sin  <p  —  2  $  cos  </>  —  <j>  +  sin  <j>  cos  0) 
—  i  V  r3  sina  <p  +  i  P  r3  (sin2  <j>  +  2  sin  /3  sin  ^>  —  2  ^  sin  /3  cos  <J>) 

+  K  T*2  (M0  —  Hr  —  P  r  sin  /3)  $  cos  0  +  A  r  cos  0  +  B. 
E I A  x'  =  —  Mo  r2  (sin  <J>— <j>  cos  0)  4- £  H  r3  (2  sin  <f>— 2  <£  cos  0  —  <£+  sin  4>  cos  ^>) 

—  i  V  r3  sin2  0  +  AC  r2  (M0  —  H  r)  <p  cos  ^  4-  A'  r  cos  0  +  B'. 
For  *  =  /3,   A  a;  =  A  #',   hence 

B-B'  =  -iPr(2  +  sin2/3)     ....      (IV.) 

Further,  for  $  =  a,    A*  =  0,    and  for  </>  =  —  a,  A  x'  =  0.     Hence,  by  add- 
ing and  subtracting, 

B  +  B'  =  Vr  (2  -  sin2  a)  -  £Pr  (2  -  sin2  a  +  2  sin  a  sin  /3) . .  (V.) 
2  Mo  (sin  a  —  a  COS  a)  —  HT*  (2  sin  a  —  2  a  COS  a  —  a  +  sin  a  COS  a) 

+  iPr  [2  -  sin2  a  +  sin2  3  -  2  cos  (a  -  /3)  +  2  (a  -  0)  cos  a  sin  /3j 
-  K  f"2(M0  -  Hr  a  cos  a  -  Pr(a  -  0)  cos  a  sin  /3J  =  0.  .(VI) 


CHAP.  IV.] 


SUPPLEMENT   TO    CHAP.  XIV. 


293 


Multiplying  HI.  by  cos  a,  and  adding  to  VI.,  we  haye 
2  Mo  sin  a  —  H  r  (2  sin  a  —  sin  a  cos  a  —  a)  +  |  P  r  (sin  a  —  sin  /3)2  =0. 
In  similar  manner,  we  have  from  eq.  (14),  Art.  2,  for  the  vertical  dis- 
placement 
E  I  A  y  =  MO  r2  (cos  4>  +  </>  sin  #)  —  i  H  r3  (2  cos  <f>  +  2  $  sin  <f>  —  sin2  0) 

^P?>3(4>  +  sin4>cos4>4-2sin/3cos4>  +  2  <£  sin  /3  sin  <?>) 
(Mo  —  H  r  —  Pr  sin  /3)  <£  sin  0  4-  Ar  sin  <J>  +  O. 
2  (cos  #  +  4>  sin  0)  —  |Hrs  (2  cos<|>  -+-  2  <£  sin  <£  —  sin8  <j>) 
+  £  V  r3  (<£  +  sin  <£  cos  <p)  +  K  r2  (M0  —  H  r)  <£  sin  ^>  +  A' r  sin  #  +  C'. 
For  $  =  0,   A  y  =  A  y',   hence 

C-C'^Pr'O  -f  sin /3  cos  0)   .     .     .     .     (VIH.) 
Finally,  f or  ^»  =  a,  Ay  =  Q.     For  $  =  —  a,   Ay'  =  0,  and  hence 

C  +  O'  =  —  SMor2  (cos  a  -I-  a  sina)  +  Hr3  (2  cos  a  +  2a  sin  a  —  sin2  a) 

+  £  P  r3 1  a  +  sin  a  cos  a  —  2  sin  (a  -  -  j8)  +  2  (a  —  3)  sin  a  sin  3| 

-  2Hr2(M0-Hr)asina-|-KPrJ(a-/3)sinasin/3   .  .  .(IX.) 
Vr  (a  —  sinaCOSa)  =  £Pr(a— 13— sina  cos  a— sin /3  cos /3 4- 2  cos  a  sin /3).  .(X.) 
18.  Determination  of  H,  V  and    M0. 
(a)   Vertical  ^Reaction. 
The  vertical  force  V  is  given  directly  by  eq.  X.     Thus 


_a  —  /3  —  sina  cos  a  —  sin  /3  cos  /3  +  2  cos  a  sin  /3 

2  (a  —  sin  a  cos  a) 
an  expression  independent  of  AC. 
Transforming  by  means  of  series,  we  have,  approximately, 


For  the  semi-circle 


TT  —  2  3  —  2  sin  (3  cos  3 

27T 


(43) 

(44) 
(45) 


From  the  exact  formula  (43)  we  have  the  following  table : 


3 

a  =  0 

a  =  10° 

a  =  20° 

a  =  30° 

a  =  40° 

a  =  50° 

o  =  60° 

a  =  90° 

0 

0.5 

0.5 

0.5 

0.5 

0.5 

0.5 

0.5 

0.5 

0.2 

0.3520 

0.3515 

0.3500 

0.3475 

0.3439 

0.3392 

0.3332 

0.3065 

0.4 

0.2160 

0.2152 

0.2130 

0.2092 

0.20B7 

0.1966 

0.1876 

0.1486 

0.6 

0.1040 

0.1033 

0.1014 

0.0981 

0.0934 

0.0874 

0.0799 

0.0486 

0.8 

0.0280 

0.0277 

0.0269 

0.0255 

0.0238 

0.0211 

0.0182 

0.0065 

1 

0 

0 

0 

0 

0 

0 

0 

0 

.a 

.P 

.  —      .  

294 


SUPPLEMENT  TO   CHAP.  XIV. 


[CHAP.  iv. 


(5)  Horizontal  thrust. 

Eliminating  M0  from  HI.  and  VI.  we  obtain,  after  reduction, 

2  sin  a  Fcos  ft  -  cos  a  +  (1  +  *)  /3  sin  /3~|  -  (1  +  *)  a  (sin2  a  +  sin2  /3) 

H  =  P L= — J .  .  (46) 

2  I  (1  +  K)  a  (a  +  sin  a  cos  a)  —  2  sin2  al 

If  we  put  Ai  =  2  sin  a  (cos  /3  —  cos  a  +  /3  sin  /3)  —  a  (sin2  a  +  sin2  /3), 
A2  =  a  (sin2  a  +  sin2  /3)  —  2  /3  sin  a  sin  /3,  , 

Bi  =  2  a  (a  +  sin  a  cos  a)  —  4  sin2  a, 
B2  =  2  a  (a  +  sin  a  COS  a), 

A  ip  A 

and  let  A  =  -^-2-,   B  =  =*-,   and  H0  =  =•*  P,  we  have 

AI  J3l  JD! 

.   1-  AK 


°  1  +  B  *' 

where  H0  is  the  value  of  H  from  the  above  formula,  when  terms  containing 
K  are  disregarded. 
Transforming  by  series,  we  have 


H  = 


From  the  exact  formula  (46)  above,  we  have  the  following  tables  : 
TABLE  FOB  H0. 


/B 

a=0 

a  =  10° 

a  =20° 

a  =  30° 

n  =  40° 

a  =50° 

a  =  60° 

a  =90° 

0 

0.4688 

0.4687 

0.4683 

0.4678 

0.4671 

0.4661 

0.4610 

0.4592 

0.2 

0.4320 

0.4317 

0.4309 

0.4291 

0.4272 

0.4243 

0.4173 

0.4017 

0.4 

0.3308 

0.3301 

0.3281 

0.3244 

0.3196 

0.3128 

0.3012 

0.2601 

0.6 

0.1920 

0.1912 

0.1887 

0.1845 

0.1784 

0.1703 

0.1578 

0.1087 

0.8 

0.0608 

0.0603 

0.0590 

0.0566 

0.0534 

0.0490 

0.0421 

0.0181 

1 

0 

0 

0 

0 

0 

0 

0 

0 

.a 

| 

CHAP.  IV.] 


SUPPLEMENT   TO    CHAP.  XIV. 


295 


For  the  values  of  the  quantities  A  and  B  we  have  the  f ollowing  table  : 
VALUES  OF  A  AND  B. 


* 

.  =  0 

a  =10° 

a  =20° 

«=30° 

a  =40° 

a=50° 

a=60° 

o=90° 

0 

3.00 

3.01 

3.02 

3.06 

3.10 

3.16 

3.25 

3.66 

0.2 

3.12 

3.15 

3.16 

3.19 

3.26 

3.34 

3.45 

4.07 

0.4 

3.57 

3.58 

3.63 

3.70 

3.81 

3.79 

4.20 

5.65 

a* 

A 

0.6 

4.69 

4.70 

4.81 

4.96 

5.20 

5.57 

6.14 

10.57 

h* 

0.8 

8.33 

8.42 

8.67 

9.10 

9.78 

10.87 

12.93 

35.62 

1 

CO 

CO 

CO 

CO 

CO 

CO 

CO 

CO 

B 

.a 

2.813 

2.825 

2.861 

2.931 

3.039 

3.198 

3.417 

5.279 

a* 

From  the  above  tables  it  is  easy  to  find  the  thrust  for  any  given  position  of 
load,  and  any  given  span  and  rise.  The  preceding  table  gives  the  reac- 
tion ;  it  only  remains  to  determine  the  moment  M0  at  crown. 

(c)  Moment  at  crown. 

From  VII.  we  have 

(sin  a  —  sin  B)2 
v 


/~  a    \ 

Mo  =  JHr  [2  —  cos  a-  -  -  )- 

\  sin  a/ 


sin  a 


Substituting  the  value  of  H,  already  given,  eq.  (46),  we  obtain 
2  Mo  f"(l  +  K)  a  (a  +  sin  a  cos  a)  —  2  sin2  a~| 

=  Pr  I  sina—  sina  cos  (a—  £)+2  sina  (cos  #—  cosa)—  sina  (sina—  sin  /3)  sin  ,8 

—  a  (cos  /3  —  cos  a)  —  (1  +  K)  {  a  (sin2  a  -t-  sin2  [3)  —  2  ^  sin  a  sin  /3  J 

+  (1  +  K)  (a  —  /3)  (a  +  sin  a  cos  a)  sin  /3  |  . 

In  similar  manner,  as  before,  for  H  we  have 

0* 


MO=MO 


l+B/c' 


where  M00  is  the  value  of  M0  when  terms  containing  Tc  are  disregarded, 
and  B  has  the  same  value  as  above.     By  series  we  have 


[~3a2- 
L 


10a/3-3/32 


and 


^4  (3a4  +  6a*  0+  45  a2  /32  +  308a/33  +  154/34)!, 


0  = 


360 


a2  (3a2  -10a/3-5/32) 


296  SUPPLEMENT   TO    CHAP.  XIV.  [CHAP.  IV. 

From  the  exact  formula  above  we  have  the  following  tables. 
VALUE  OF  M00. 


& 

a  =  0 

0=10° 

a  =  20° 

o  =  30° 

a  =40° 

a  =50° 

a  =  60° 

a  =  90° 

0 

0.09375 

0.09426 

0.09582 

0.09849 

0.10241 

0.10777 

0.11613 

0.15108 

0.2 
0.4 

0.01606 
0.02025 

0.01615 
0.02021 

0.01658 
0.02008 

0.01742 
0.01973 

0.01846 
0.01942 

0.02063 
0.01888 

0.02280 
+ 
0.01719 

0.03083 
+ 
0.01441 

0.6 

0.02400 

0.02393 

0.02369 

0.02326 

0.02267 

0.02181 

0.02001 

0.01454 

0.8 

0.01025 

0.01019 

0.00999 

0.00962 

0.00917 

0.00850 

0.00716 

0.00332 

1 

0 

0 

0 

0 

0 

0 

0 

0 

.a 

.Pa 

VALUES  OF  B  AND  C. 


• 

a=0 

a  =  10° 

a  =  20° 

a  =  30° 

a  =40° 

a  =  50° 

a  =60° 

o  =  90° 

0 

7.50 

7.52 

7.57 

7.78 

7.94 

8.52 

8.61 

11.12 

0.2 

28.13 

28.09 

27.99 

28.14 

28.43 

27.99 

27.76 

31.81 

0 

0.4 

12.50 

12.61 

13.05 

14*08 

15.40 

17.35 

21.11 

37.47 

a4 

0.6 

4.69 

4.77 

5.00 

5.51 

6.24 

7.31 

9.10 

20.27 

0.8 

2.74 

2.85 

3.20 

3.89 

4.94 

6.61 

9.77 

42.53 

1 

1.88 

00 

00 

00 

00 

00 

00 

00 

B 

.a 

+2.81 

+2.83 

+2.86 

+2.93 

+3.04 

+3.20 

+3.42 

+5.28 

a4 

Here,  as  always,  a  negative  moment  denotes  tension  in  lower  or  inner 
flange.  We  see  at  once  from  the  table  that  the  maximum  compression  in 
this  flange  at  crown  does  not  occur  for  full  load,  but  for  load  extending 
from  both  ends  towards  the  crown  as  far  as  about  gths  of  the  span  or  fths 
of  the  half  span.  "Within  the  middle  half  of  the  arch,  then,  a  load  any- 
where causes  tension  in  lower  flange  at  crown— outside  of  this  middle  half 
a  load  anywhere  causes  compression  in  the  lower  flange  at  crown.  For 
large  central  angles,  K  may  be  disregarded,  and  we  have  simply  M  =  M00. 


CHAP.  IV.]          SUPPLEMENT  TO  CHAP.  XIV.  297 

19.  Intersection  Curve.— From  Art.  13  we  have 
V  r  sin  3  -  M0 

y=       — ^ —  — • 

Hence,  by  substitution  of  the  values  of  V,  H  and  M0, 

-  rj£  l~l  -  12°(a*  -2afl-/32)  1 
y  ""    10  L    "  a4  (a  +  ft)2  *J 


which  is  the  equation  given  in  the  text,  for  which  a  table  is  there  given. 

2O.  Direction  Segments.— It  will  in  the  present  case  be  found 
most  convenient  to  determine  the  directions  of  the  resultants  by  d  and  e2 
equation  (34). 

M!  M2       - 

Thus,  Ci  =  -  -|p      c2  =  —  -jj-. 

But  M!  =  Mo  -  H  h  +  (P  —  V)  a  —  P  z,   M2  =  M0  —  H  h  +  V  a. 
We  have,  by  series,  then  the  approximate  formulae, 

45Ififl  2h     T  451 


15  («+«)  L  A7i2  J  15  («-+- 

where  positive  values  of  Ci  and  c2  are  laid  off  upward  above,  negative 
values  downward  below,  the  centres  of  gravity  of  the  end  cross- sections. 

From  the  preceding  tables  we  can  calculate  easily  in  any  case  H  and  V 
an^.  Mo,  and  thus  check  the  results  obtained  by  the  method  of  Chap.  XIV. 
The  formulae  above  for  d  and  c2  do  not  admit  of  tables,  nor,  in  fact,  are 
such  needed.  They  are  sufficiently  simple  for  ready  insertion. 

Thus,  by  the  aid  of  our  tables,  having  computed  V  and  H,  and,  if  neces- 
sary, Mo  and  0o,  we  can  by  the  method  of  moments,  as  explained  in  Chap. 
XIV.,  Art.  162,  readily  calculate  the  strains  in  the  braced  arch,  whether 
continuous  at  crown  and  fixed  or  hinged  at  the  ends,  or  hinged  at  both  ends 
and  crown. 

2O  (ft).  Transformation  Series.— We  have  in  the  preceding  repeat- 
edly made  use  of  series  in  the  transformation  of  angular  functions,  such  as 
sin,  cos,  etc.,  into  functions  of  the  arc  itself.  We  group  here,  for  conve- 
nience of  reference,  the  series  thus  used : 


sin  x  =  x  (I  -  i»2  +  7^0  aj4  -  <roL4-u  «6  +  Wainr  &  -  uinrunn)  o  * 

COS  X  =  1  -  i  X2   +  &  X*  -  Tib  «?  +  Toku  «"  ~  TSjtv 0  0  ^  ° 

sin  2x  =  x2  (I  -  -U2 
cos2  0=1-  x 
sin  x  cos  x  =  x  (I  —  I  x2  -t-  -ft  x*  —  3-73-  x°  +  dW  «9  —  ITT&TS  a?1  °  +  ...) 


298  SUPPLEMENT   TO   CHAP.  XIV.  [CHAP.  IV. 

COS3  X  =  1  -  t  X9  +  U4  -  #jj  Xs  +  Tfffcj  X8  -  TTWoo  Z1  °  +  ... 

tan  a;  =  x  (1  +  1  x2  +  -fr  a*  +  &  x«  +  7f|F  zs  +  rHfle  «10  4-  ...) 


1       # 

cot  *  =  -  -      (1  +  -iV  x2  +  ^  tf  +  -nky  a;6  +  ^if  «?  «8  +  ...) 


sin  x  siny  =  xy  [l 
+  7  32 


cos  a;  cos  y=l- 
+15 


sin  «  cos  y=a  [l- 
+35 


CHAP.  V.]          SUPPLEMENT  TO  CHAP.  XIV.  299 


CHAPTER    V. 

INFLUENCE   OF   TEMPERATURE. 

21.  Oeneral  considerations.—  When  the  temperature  of  a  per- 
fectly free  body,  which  possesses  in  every  direction  the  same  coefficient  of 
elasticity  and  expansion,  changes  equally  at  all  points,  there  can  be  no 
strains  in  the  body.  For  were  there  such  strains,  then,  as  there  are  no 
outer  forces,  there  could  be  no  equilibrium. 

If,  however,  the  change  of  temperature  is  not  the  same  at  all  points  ;  or 
if  the  body  is  not  free,  so  that  it  is  possible  for  outer  forces  to  act,  there 
are  strains. 

In  the  following  we  assume  the  change  of  temperature  to  be  everywhere 
the  same,  but  that  the  body  is  not  free. 

We  assume  that  at  a  certain  temperature  t0  no  strain  exists  in  the  body, 
and  call  this  the  mean  temperature.  The  deviation  above  or  below  the 
mean  temperature  we  call  +  t  or  —  t,  and  denote  the  coefficient  of  expan- 
sion for  one  degree  by  e. 

The  determination  of  the  strain  in  a  straight  beam  held  at  both  ends,  is 
very  simple.  If  the  length  is  I,  its  relative  change  of  length  is  e  t.  Since, 
however,  it  cannot  expand,  the  strain  S  per  unit  of  area  is  precisely  as  great 
as  the  force  which  would  be  required  to  produce  this  relative  elongation, 

or  from  eq.  (4)  s  =  +  E  e  t    ........     (48) 

If  the  area  of  cross-section  is  A,  then  the  strain  at  each  end  is 


In  equation  (48)  it  is  assumed  that  a  compressive  strain,  due  to  +  t,  is 
positive,  a  tensile  strain,  due  to  —  t,  is  then  negative. 

22.  Influence  of  Temperature  on  the  Arch.—  Since  by  a 
change  of  temperature  the  length  of  the  arch  varies,  while  the  span  remains 
always  the  same,  the  shape  or  curvature  must  change,  which  naturally  must 
give  rise  to  strains  and  outer  forces.  In  the  following  we  have  only  to 
determine  these  outer  forces,  since,  as  shown  in  Chap.  XIV.,  these  are  all  we 
need  to  determine  the  strains  themselves. 

*f        /  TVT\ 

The  relative  change  of  length  is  from  eq.  (8),  Art.  I.,  ^-^  JG  +  —  J. 


This  change  is  caused  by  the  outer  forces.     The  relative  change  of  length 
due  to  temperature  alone  is  e  t.    Hence  the  total  relative  change  of  length  is 

Gr  +  M 


300  SUPPLEMENT   TO    CHAP.  XIV.  [CHAP.  V. 

Hence  the  change  of  length  of  the  axis  is 

—    /    d  s  —  f  t  s (50) 

The  change  of  the  angle  between  two  infinitely  near  cross-sections,  and  the 
actual  turning  of  a  cross-section  is  from  (9)  and  (12),  Art.  1,  given  by 

M       G  r  +  M 


d  0  -  El         EA?> 

Gr 


1    /»M  1     /^ 

l-(j>  =  Ej  Tds+<B~rJ 


A 

Finally,  from  (14)  we  have 

d  s 


ds 


/(*&d  s 
&<!>dy  +   I  -j^ 

/ 
^tX 


Substitute  in  these  last  two  equations  for  A  <j>  and  -  their  values  from 

(.1  8 

(49)  and  (52).     The  double  integral  thus  arising  can  be  resolved  by  par- 
tial integration. 

Thus  /*  //(*)<& 

Applying  this,  we  obtain 

r  r& 

Ax—  —  yA0+  /  yd&+  I  —  —  dx 

J  J    *'  ....     (5S) 

/r&ds 
zdA<f>+  I  -j—dy 

We  shall  assume  in  the  following  the  axis  always  circular. 

23.  Fundamental  Equations  —  General.  —  Upon  this  assump- 
tion of  a  circular  axis  we  have  generally 

G  =  H  cos  <£,    N  =  H  sin  <f>  "J 

M  =  M0  -f  Hr(l-cos^)   I      .....     (54) 

Gr  +  M  =  M0  +  Hr  j 

Hence,  from  the  preceding  Art, 

H  r  +  Mo  '  /** 
A  s  =  —  —  —   -Id  +  —Ttt$    .....     (55) 

E  A      Jo 


AEy.  ° 


A  a?  =  —  r  A  0  (1  —  cos  <j>)  H /  M  (1  —  cos  $>)  d  $ 


r  +  Mo  r     _ 

EA      J      *      *««?*+ 


CHAP.  V.]  SUPPLEMENT  TO  CHAP.  XIV.  301 

24.  Arch  with  three  Hinges.— If  there  are  three  hinges,  the  mo 
ment  M0  at  the  crown  must  be  zero,  and  therefore  M  =  H?*(l  —  cos  <£). 
But  for  0  =  a,  M  must  also  be  zero,  hence  H  r  (1  —  cos  a)  =  0,  and  therefore 
H  is  zero.     Then  for  any  point  G  =  0,   and  M  =  0,  and  N  =  0.     That  is, 
for  the  arch  with  three  hinges  there  are  for  a  change  of  temperature  no  outer 
forces,  and  hence  no  strains. 

25.  Arch  hinged  at  Ends.— Here,  since  f or  0  =  a,   M  =  0,  we 
have  from.  (54) 

Mo  =  —  H  r  (1  —  cos  a)        \ 

M  =  -  H  r  (cos  0  -  cos  a)  I (58) 

G  r  ~\-  M  =  +  H  r  cos  a       J 
.  From  (56),  since  for  9  =  0,     A  0  =  0, 

A  ^  =  — /   (cos  0  —  cos  a)  d  0  +      cos  a    /  d  $  .  .  (59) 

J  o  Jo 

From  (57),  since  for  0  =  0,    A  *  =  0, 

Hr'f  f*  C*  1 

A  02=  I    (1  —  COS  0)      /    (COS  0  —  COSa)^0—      /     (1—  COS  0)  (COS  0  —  COS  a)  d  0    I 

L  Jo  Jo  J 

H  r  cos  a  cos  61, 

•I ^r~a I   d  ^  —  r  e  t  sin  (j>. 

E  A  / 

Jo 

For  ^  =  a,  this  becomes  zero,  and  we  have  for  the  horizontal  thrust 
_,  _  E  c  t  sin  o 

r2     Ca  >cos«    /""  cos2  a    /*"    ' 

—     I   cos  <4  (cos  0— cos  a)  d <h —     I   (cos0— cos  a)d6-\ i —     I   a$ 

I    /  I        /    ^  A      / 

«/0  i/0  t/0 

....     (60) 
Performing  the  integrations  indicated  (Art.  7),  and  putting,  for  brevity, 

K  = ,  we  have 

Ar2 

„_  2EIf£sina 

r2  (a  —  3  sin  a  cos  a  +  2  a  cos'2  a)  +  2  x?*2  a  cos2  a 

By  series  (Art.  20),  we  obtain  the  approximate  formula 


r2  (2«4  +  15/c)      8AA2  +  151 

The  above  are  the  expressions  given  in  Art.  165  without  proof.  The  less 
h,  the  greater  for  equal  dimensions  is  H.  For  h  =  0,  we  have 
H  =  E  A  e  t,  as  we  should  have  for  a  straight  beam. 

26.  Arch  without  Hinges.— In  this  case  we  have  the  general  equa- 
tions (54)  and  (55),  which  apply  directly  without  change. 

From  (56),  since  for  0  =  0,   A  0  =  0,  we  have 


302  SUPPLEMENT   TO    CHAP.  XIV.  [CHAP.  V. 

From  (57),  since  for  $  =  0,    A  x  —  0,  we  have,  inserting  the  value  of  A  <£, 
above, 

«."r    A    r  ,  i 

A  x  = 1  cos  &    I    dfy  —     I   cos  0  a  0  I 

H   :JF  A  A 

~^F  I  C1  ~  cos  <?)    /  (1  -  cos  0)  <Z0  -    /  (1  - 
EIL  Jo  Jo 

r* 

»    /  d  $  — 
Jo 


-+-  — =-. - — -  cos  0    /  d  $  —  r  1 1  sin 
E  A 

For  0  =  a,   A  ^  =  0,   hence  from  the  first  of  these  expressions 
r2    C"  •       1     Ca 


= 


r*  c*    i  r 

—    I   dQ  +  j    Ic 

ljo  AJo 


If  the  distance  at  which  the  horizontal  thrust  H  acts  from  the  crown  is 
«o,  we  have  M0  =  —  H  e0,  whence  we  see  at  once  that  e0  is  the  fraction  in 
(63)  multiplied  by  r.  For  </>  =  0,  A  a  must  also  be  zero,  and  we  thus 
obtain  another  relation  between  Mo  and  H  which  does  not  contain  A.  If 
we  multiply  the  expression  thus  obtained  by  r  cos  a,  and  then  subtract  the 
result  from  that  previously  obtained  for  0  =  0,  A  $  =  0,  we  have 


^    /"a  /*a 

|      /  (l-cos0)d0-    /  (l- 


....     (64) 
Performing  the  integrations  indicated  in  (63),  we  have  (Art.  7) 


where,  as  before,  K  =  -  . 
Ar2 

From  (64)  we  obtain 

M0  r  sin  a  —  iHr2  (a  —  2  sin  a  +  sin  a  cos  a)  =  —  E  I  f  t  sin  a. 
Inserting  the  value  of  M0  above,  we  have 

H  =  —  8BI,<(l  +  ,).rin. 

2  [(1  +  K)  (a2  +  a  sin  a  COS  a)  —  2  sin2  a] 


r 
and  hence 


M  —  —  - 

""»•[(!  +  f)  (a2  +  a  sin  a  cos  a)  —  2  sin2  a] 

From  these  two  we  obtain  for  the  point  of  application  of  H 

«  =  -igt  =  +  (1+')''-rin"r     ....     (88) 
(1  +  K)  a 


CHAP.  V.]          SUPPLEMENT  TO  CHAP.  XIV.  303 

By  series,  we  hare  (Art.  20)       , 


without  reference  to  K,  e0  =  $  h. 

For  small  central  angles,  then,  for  which  <  may  be  disregarded,  the 
thrust  given  above  by  (66)  acts  at  }  Ji  below  the  crown  for  a  rise  of  tem- 
perature of  t  degrees  above  the  mean.  For  a  decrease  of  temperature  be- 
low the  mean  it  acts  above,  M0  is  negative,  and  the  strain  in  the  lower 
flange  tensile. 

Further,  we  have,  by  series,  the  approximate  formulae 


r2(a4  +  45*)      4  A  A2  +451 


These  are  the  expressions  given  in  Art.  165  without  proof. 


304  SUPPLEMENT   TO   CHAP.  XIV.  [CHAP.  VI. 


CHAPTER  VI. 

PAETIAL     UNIFORM     LOADING. 

27.  Notation.  —  In  the  preceding  discussion  of  the  -arch  we  have  con- 
sidered the  influence  of  a>  single  concentrated  load  only,  and  this,  as  we 
have  repeatedly  seen  in  the  case  of  the  simple  and  continuous  girder,  etc., 
is  sufficient  for  full  and  accurate  solution.     When  once  we  are  able  to  find 
and  tabulate  the  strains  in  every  piece  due  to  a  single  load  in  any  position, 
the  thorough  solution  becomes  simply  a  question  of  time. 

It  may  often  happen,  however,  that  we  may  wish  to  determine  the  strains 
for  a  full  load  only,  or  for  a  uniformly  distributed  load  extending  from 
one  end  to  some  given  point.  In  such  case  it  would  be  unnecessarily  tedi- 
ous to  obtain  our  result  by  the  successive  determination  and  addition  of  all 
the  intermediate  apex  loads.  We  may  easily  deduce  from  the  preceding 
the  general  formulae  for  partial  loading  also. 

As  before,  we  shall  let  a  =  the  half  span,  h  =  the  rise,  I  the  moment  of 
inertia,  and  A  the  area  of  the  cross-section.  But  we  shall  represent  by  p 
the  load  per  unit  of  length  of  horizontal  projection,  and  by  z  the  distance 
of  the  end  of  the  load  extending  from  the  left,  from  the  crown.  This  dis- 
tance z,  from  the  crown  to  the  end  of  load,  is  then  positive  towards  the 
left.  In  the  circular  arch  the  angle  subtended  by  this  distance  -z  we  call  /3. 
The  angle  /3  is  then  positive  to  the  left  of  the  vertical.  The  angle  sub- 
tended by  the  half  span  is,  as  before,  a.  For  /3  =  a,  then,  or  for  z  —  a, 
there  is  no  load  upon  the  span.  For  /3  =  0,  or  z  =  0,  the  load  extends 
from  the  left  to  the  centre.  For  /3.=  —  a,  or  z  =  —  a,  the  load  covers  the 
whole  span.  Pis.  23  and  24,  Figs.  91  and  92,  still  hold  good,  therefore,  for 
our  notation.  We  have  only  to  conceive,  instead  of  the  concentrated  load 
P,  a  uniformly  distributed  load,  per  horizontal  unit,  extending  from  left  end 
as  far  as  the  position  of  P.  This  much  being  premised  as  to  notation,  we 
shall  treat,  as  before,  the  three  cases  of  arch  hinged  at  crown  and  ends, 
hinged  at  ends  only,  and  without  hinges. 

A.    ARCH  HINGED  AT   CROWN   AND  ENDS. 

28.  Reaction.  —  This  case  is  too  simple  to  demand  any  extended 
notice,  in  view  of  what  has  already  been  said.    We  have  from  eq.  (16),  Art. 
3,  for  the  reaction  at  the  left  or  loaded  end,  for  concentrated  load, 


2a 
If  now  we  put  P  =  p  d  z,  and  integrate,  we  have 

r     .,    a  +  z         2az  +  z 
V  =  /  p  d  z  —  —  =  p  —  -.  --  h  C, 
J  *          2a  ±a 

where  O  is  the  constant  of  integration.     By  taking  the  proper  limits,  we 
can  eliminate  this  constant,  and  thus  obtain  the  reaction  for  load  covering 


CHAP.  VI.]  SUPPLEMENT   TO    CHAP.  XIV.  305 

any  desired  portion  of  the  span.  As  we  shall  in  every  case  suppose  the 
load  to  extend  from  the  left  end  up  to  any  point,  we  shall  take  the  limits 
of  z  =  a  and  2,  and  therefore  obtain 


For  z  —  a,  this  is  zero,  as  it  should  be,  since  then  the  load  has  not  come 
on.  For  z  =  —  a,  the  load  extends  over  the  whole  span,  and  V  =  p  a,  or 
half  the  whole  load,  as  it  should.  We  might  have  obtained  this  result  at 
once  by  moments.  Thus, 


V  x  3a  =      a  - 


but  have  preferred  the  above  method  as  showing  how  uniform  loading  is 
deduced  directly  from  concentrated  by  inserting  pdz  for  P  and  inte- 
grating. 
29.  Horizontal  Thru§t.  —  In  precisely  similar  manner  we  have 

from  (21),  Art.  4,  for  the  thrust  due  to  concentrated  load  P,  H  =  P  ^  ~  e\ 

2  h 

Put  P  —  p  d  3  and  inte-grate  between  the  limits  z  —  a,  and  2,  and  we  have 


For  z  =  a,  this  is  zero,  as  should  be.     For  z  =  —  a,  or  for  full  load  over 

'D  QJ^" 

whole  span,  H  =  ^T.     We  may  also  deduce  the  above  equation  directly 

tii  fi 

by  moments. 

The  above  formulse  (71)  and  (72)  are  all  that  we  need  either  for  calcula- 
tion or  diagram.  They  apply  evidently  equally  well,  whether  the  arch  be 
circular  or  parabolic,  or,  in  general,  whatever  its  shape  may  be.  The  form 
has  no  influence  upon  either  the  thrust  or  the  reaction. 

For  the  moment  at  any  point  whatever,  whose  distance  horizontally 
from  crown  is  x  and  vertically  below  crown  y,  we  have  at  once 

M  =  H  (h  -  y)  -  V  (a  -  x)  +  2-  (a  -  a)2. 

2 

If  this  point  is  an  apex,  then  the  moment  divided  by  depth  of  arch  at 
this  point  is  the  strain  in  flange  opposite  that  apex.  A  positive  moment 
throughout  this  work  always  indicates  compression  in  the  inner  or  lower 


B.  AKCH  HINGED  AT  ENDS  ONLY. 

30.  Reaction.—  The  vertical  reaction  at  the  end  is  precisely  the  same 
as  before  for  three  hinges,  and  is  given  by  equation  (71).     This  reaction  is 
evidently  independent  of  the  shape  of  the  arch,  and  the  above  formulae 
holds  good  generally. 

31.  Horizontal    Thr«§t—  Parabolic    Arch.—  We  must  here 
distinguish  the  shape  of  the  arch,  and  treat  first  the  parabola.     We  have 
already  from  eq.  (27),  Chapter  III.,  Art.  8,  for  a  single  load, 


5 

~ 


20 


306  SUPPLEMENT   TO   CHAP.  XIV.  [CHAP.  VI. 

We  have,  as  before,  simply  to  make  P  =  p  d  z,  and  then  integrate  between 
the  limits  z  =  a  and  z  indeterminate. 
We  thus  find  at  once 

n      r  -i 

.     .      (73) 

For  z  =  a,  this  reduces  to  zero,  as  it  should.    For  z  =  —  a,  the  load  covers 

the  whole  span,  and  we  have  H  —^-.      For  z  =  o,  the  load  reaches  from 

2  h 

2 

the  left  as  far  as  the  crown,  and  H  —-r-^-.    The  formulae  is  simple,  and  re- 

4ft 

quires  no  table.     Numerical  values  may  be  easily  inserted. 

32.  Horizontal  Thrust— Circular  Arch.— As  already  noticed, 
the  vertical  end  reaction  for  this  case  has  been  given  in  eq.  (71).  It  re- 
mains to  determine  the  thrust.  We  have,  as  before,  simply  to  insert  pdx  = 
p  r  cos  /3  d  3  in  place  of  P  in  the  expression  for  the  thrust  for  concentrated 
load  of  Art.  10,  and  then  integrate  between  the  limits  /3  =  a  and  /3  inde- 
terminate. 

We  have  thus  similarly  to  that  Art. 


where  H0  is  the  value  of  H  when  terms  containing  K  are  neglected,  or 

H.=?;|-;andA  =  ^B  =  B1 
12  BI  AI  BI 

The  quantities  Ai,  Bi,  A2  and  B2  are  as  follows : 

A,  =  7  sin3  a  +  3  a  cos  a  —  3  sin  a  —  6  a  cos  a  sin2  a  —  6  sin2  a  sin  /3 
+  2  sin3  /3  —  3  /3  cos  a  —  9  cos  a  sin  £  cos  /3  +  12  cos2  a  sin  /3 
+  12  a  cos  a  sin  a  sin  £  —  6  £  cos  a  sin2  /3. 
A2  =  3  [2  a  cos  a  sin2  a  +  a  cos  a  —  sin  a  cos2  a  —  4  a  cos  a  sin  a  sin  j3 

+  2  |3  cos  a  sin2  0  —  /3  cos  a  +  cos  a  sm  £  cos  3]. 
Bj  =  a  —  3  sin  a  cos  a  +  2  a  cos2  a.         B2  =  2  a  cos2  a. 

These  expressions  can  be  tabulated  as  in  Art.  10,  or  developed  into 
series  as  in  that  Art.,  and  the  formula  thus  made  practically  available. 

For  £  =  a,  we  have  H  zero,  as  should  be.     For  ]3  =  —  a,  we  have  the 
load  covering  the  entire  span. 
For  this  case  we  have 

H_l     f  sin3  q  -  3  (1—2  sin2  a)  (sin  a  -acos  a)  -  3  K  cos  a  (a  +  2  a  sin»  a  —  sin  a  coa  a) 
'  ~  6  "a  •+•  2  a  cos'-i  a  —  3  sm  a  cos  a  +  2  K  a  cos1^  a 

For  the  SEMI-CIRCLE,  this  reduces  simply  to 

H  =  —  pr  =  0.424  pr. 

3  7T 

In  any  case  where  exact  results  are  desired,  eq.  (74)  must  be  used,  and  a 
table  calculated  for  the  central  angle  a.  We  have  approximately  by 
series  also,  more  especially  for  small  central  angles,  or  for  a  large  in  respect 
to  hj  for  total  load  over  whole  span : 

j>a2        8  Aft*         _pa*         8ft2  „ 

2  ft  15  I  +  8  A  ft2  ~  2  ft  15  g*  +  8  ft2 


CHAP.  VI.]  SUPPLEMENT   TO    CHAP.  XIV.  307 

where  A  is  the  area  and  I  the  moment  of  inertia  of  cross-section,  and  g 
the  radius  of  gyration.  In  framed  arches  this  may  be  taken  as  approxi- 
mately equal  to  the  half  depth  from  centre  to  centre  of  flanges. 

C.    ARCH   WITHOUT   HINGES  —  FIXED   AT   ENDS,    CONTINUOUS   AT   GROWTH. 

33.  Parabolic  Arch  —  Formulae  for  V,  H  and  M.  —  In  this 
case  the  reactions  no  longer  follow  the  law  of  the  lever,  and  eq.  (71),  there- 
fore, no  longer  holds  good. 

(a)   Vertical  reaction  at  unloaded  end. 

We  have  from  eq.  (28),  Art.  14,  for  the  reaction  at  the  right  end  for  a 
single  load, 

1       (a  -  zY  (2  a  +  e) 
~4P          ~~^~~ 

Making  P  =pdz,  and  integrating  between  the  limits  z  =  a  and  a,  we 
find  the  reaction  for  a  load  coming  on  from  left, 


V  = 


L.    3  a4  -  8  a:J  z  +  6  a2  z2  -  z*\     .     .     .    (76) 


for  a  full  load  z  =  —  a  and  V=pa,  as  should  be. 

(5)  Horizontal  thrust. 

In  like  manner  we  have  for  the  horizontal  thrust  at  end  from  (36),  Art. 
14, 

15  p  g*  -  2  a2  z*  +  g* 
H  =  32  a*h 

Replacing  P  by  p  d  z,  and  integrating  as  before,  we  obtain  directly 


H=  -—  -15a«2  +  Wa*z*-3z*       .     .     (77) 

in  $2 

for  a  full  load  z  =  —  a,  and   H  =  £—  —  . 

to  h 

(c)  Moment  at  unloaded  end. 

In  precisely  similar  manner  we  have  from  (37),  Art.  14, 

_         1       (a  —  zY  (3  a2  -  10  a  z  -  5  g2) 
~32  ~~^~~ 

Putting  P  =  p  d  z,  and  integrating,  we  harve  for  the  moment,  always  at 
the  right  end,  or  for  load  not  extending  past  the  centre,  at  crown 


M 


=  -£-  [3a4z-8a3z* +6a*z*  -  g8]       .      .     (78) 
32  a    L  J 


For  z  =  a,  this  is  zero,  as  should  be.     For  z  —  0,  or  for  load  extending  as 

far  as  crown,  it  is  also  zero.     For  z  =  —  a,  the  moment  at  the  end  is  —  ^— . 

2 

A  negative  moment,  as  always,  denotes  tension  in  lower  flange. 

Just  as  for  concentrated  load,  as  shown  in  Art.  14,  as  the  load  comes  on, 
the  moment  at  crown  is  positive,  and  increases  with  increasing  load  up  to 
a  certain  point,  beyond  which  any  load  causes  a  negative  moment,  and  be- 
yond which  the  moment  at  crown,  therefore,  decreases,  until,  when  the  load 


303 


SUPPLEMENT   TO   CHAP.  XIV. 


[CHAP.  vi. 


reaches  the  crown,  it  becomes  zero.  This  point,  which  gives  M0,  the  mo- 
ment at  crown,  a  positive  maximum,  is  at  a  distance  z  =  —  a  +  aVf  — 
0.264911  a,  or  nearly  i  a  from  the  crown. 

The  values  of  V2,  H  and  Ma  (M2  and  V?  being  always  the  moment  and 
reaction  at  unloaded  end),  for  various  values  of  z,  are  given  in  the  following 
table: 


z 

V2 

H 

M2 

z 

V2 

H 

M2 

1 

0 

0 

+ 

—0.1 

0.2349 

0.29656 

-0.01206 

0.9 

0.0002437 

0.000579 

0.0001096 

-0.2 

0.3024 

0.34128 

-0.03024 

0.8 

0.0014 

0.00421 

0.00076 

-0.3 

0.3707 

0.38242 

-0.055611 

0.7 

0.00624 

0.01643 

0.002185 

-0.4 

0.4459 

0.41846 

-0.08918 

0.6 

0.0144 

0.02889 

0.00432 

-0.5 

0.5273 

0.44824 

'-0.131836 

0.5 

0.0273 

0.051757 

0.006836   i 

-0.6 

0.6144 

0.47104 

-0.18432 

1 

0.4 

0.0459 

0.08346 

0.00918 

-0.7 

0.7062 

0.48667 

-0.24718 

0.3 

0.07074 

0.11777 

0.010611 

-0.8 

0.8014 

0.49884 

-0.32076 

0.2 

0.1024 

0.15872 

0.01024 

-0.9 

0.90024 

0.49942 

-0.405109 

0.1 

0.1412 

0.20315 

0.007062 

-1 

1.0000 

0.5000 

-0.5000 

0 

0.1875 

0.25 

0 

a 

pa 

pa* 

p  a2 

a 

pa 

p  az 

p  a2 

h 

k 

It  will  be  seen  that  the  moment  at  the  unloaded  end,  which,  as  long  as 
the  load  is  left  of  crown,  is  the  moment  at  crown  also  ;  increases  as  the 
load  passes  on,  is  positive  and  increases  up  to  about  z  =  .25  a.  Then  it 
diminishes,  becomes  zero  when  the  load  reaches  the  crown,  changes  to 
negative  as  the  load  passes  the  crown,  and  this  negative  value  increases  up 
to  full  load  when  it  is  —  i  p  a2.  For  full  load,  then,  the  lower  end  flanges 
are  in  tension.  At  the  crown  the  moment  is  zero,  and  the  compression 
there  in  both  flanges  is  due  to  H  only. 

34.  Circular  Arch— Formulae  for  V,  H  and  H. 

(a)    Vertical  Reaction. 

Here  we  have  r  sin  /3  =  x,  r  cos  j3  d  /3  =  d  $,  and  P=pdx  =  prcosfidp. 
Inserting  this  in  place  of  P  in  eq.  (43),  Art.  18,  and  integrating  between 
the  limits  /3  =  a  and  /3  indeterminate,  we  have  for  the  reaction  at  w?aloaded 
end,  or  for  reaction  at  crown  when  load  does  not  extend  past  the  crown, 

V  =  — ^L |COS    a  —  cos  a  —  a  sin  /3  +  cos  ft  +  /3  sin  |8 

2  (a  —  sin  a  cos  a)  L     3 

+  sin  a  cos  a  sin  0  -  22jl5  _  cos  a  sin2  01. . .  .(79) 

J 


CHAP.  VI.]          SUPPLEMENT  TO  CHAP.  XIV.  309 

For  0  =  a,  this  is  zero,  as  should  be,  since  then  the  load  is  not  upon  the 
span.     For  0  =  —  a,  V  =  p  r  sin  a,  as  should  be,  for  full  load  over  whole 

span.     For  the  semi-circle,  a  =  90°  =  — ,    sin  a  —  1,  cos  a  =  0,  and 

2 

TT  nr»H3    R 

—  -  sin  0  +  cos  ft  +  0  sin  0  — 


V  —  p  r  — — . 

7T 

If  the  semi-circle  is  uniformly  loaded  over  whole  span,  0  =  —  a  =  —  90° 

=  —  — ,  sin  0  =  —  1,  cos  0  —  0,    and  V  —  pr,    as  should  be.     The  for- 
2 

mula  (79)  above  is  precisely  the  same  as  that  given  by  Capt.  Eads  in  his 
Report  to  the  Illinois  and  St.  Louis  Bridge  Co.,  May,  1868. 

(5)  Horizontal  Thrust. 

In  similar  manner,  from  eq.  (46),  Art.  18,  by  inserting  p  d  x  =  p  r  cos  0  dj3 
in  place  of  P,  and  integrating  between  0  =  a  and  8,  we  have  similarly  to 
Art.  32 

H  =  Ho^A_* (80) 


and 


,  , 

12       -t»i  AI  Uj 

A!  =3  a—  3  sin  a  cos  a—  2  a  sin2  a—  3  0—  9  sin  0  cos  0+12  COS  a  sin  0 

—6  0  sin2  0+6  a  sin  a  sin  0+2  a  ^?—  -, 

sin  a 

A2=3a—  3  sinacosa+2a  sin2  a+60sin2  0—  3  0+3  sin/3  cos0 

—  6  a  sin  a  sin  0—2  a  —  -  , 
sin  a 

Bi=a  (a+sin  a  COS  a)  —2  sin2  a,         B2=a  (a+sin  a  COS  a). 
Formula  (80)  agrees  exactly  with  that  given  by  Capt.  Eads  in  the  Report 
above  quoted,  if  terms  containing  K  are  neglected.    Since  K  =  -  -,  where  I 

is  the  moment  of  inertia  and  A  is  the  area  of  cross-section  ;  r  being  the 

radius  ;  for  small  central  angle  r  is  very  large  in  proportion  to  —  ,  or  the 

A 

square  of  the  half  depth.  In  such  case,  then,  K.  may  be  neglected.  For 
0  ==  a,  we  have  H  =  0,  as  should  be.  For  0  =  —  a,  we  have  load  over 
entire  span,  and 

H_i.         •       3  a—  2  n  sin-  «—  3  sin  a  cos  a—  K  (3  a+2  a  sing  a—  3  sin  a  cos  a) 
(1+K.)  a  (a+sin  a  cos  a)  —2  sin2  a 

Approximately  we  have,  by  series,  for  full  load,  from  (80)  : 


_ 
2h  2  22 


310  SUPPLEMENT   TO   CHAP.  XIV.  [CHAP.  VI. 

For  o  =  90°,  or  for  semi-circle,  we  have  from  (80) 

t [37r-6/3+18sin/3cos/3-12/3sin2£+67rsin/3 


•hK  (-3  7T-12  ft  sin2  /3+6  £-6  sin  ft  cos  ft+Q  ir  sin  /3+2  TT  sin3  /3j. 
For  /3  =  —  90°,  or  for  full  load  upon  semi-circle, 

PL- V«,-9. 


(c)  Moment  at  unloaded  end. 

From  Art.  18  (c)  we  have,  for  concentrated  load, 

(sin  a  —  sin  /3)2 


/2- 
\ 


cosa-       -- 


sin  a/  Sill  a 


The  value  of  H  we  have  already  given  in  (80). 

Inserting  in  the  second  term  p  d  x  =  p  r  cos  ft  d  ft  for  P,  and  then  inte- 
grating, we  have 

M  =  AH-B    .......     (82) 

where         A  =  |  r  1  2  —  cos  a  --  ^—  )  =  JT-?  —  (2  —  cos  a  —  -^—  \ 
y  sin  a/       2  sin  a  \  sin  a/ 

and  B  =     P  f        (sin  a  —  sin  ft  }  . 

12  sin3  a  \  / 

For  a  uniform  load  over  whole  span,  ft  =  —  a,  and 

^    .  (83) 


sm  «/  3 

"We  have  from  (83),  by  series,  the  approximate  formula  for  moment  at 
crown 

175y«  a*  . 


where  A  is  the  area  and  I  the  moment  of  inertia  of  cross-section,  g  the 
radius  of  gyration,  or,  approximately,  the  half  depth  for  framed  arch  —  as 
always  a  negative  moment  indicates  tension  in  lower  or  inner  flange. 

Equations  (79),  (80)  and  (82)  may,  if  desired,  be  tabulated  as  in  Art.  18. 
For  small  central  angles,  or  for  h  small  with  respect  to  a,  K  may  be  disre- 
garded, and  the  results  already  given  for  parabola  (Art  33)  may  be  taken 
as  sufficiently  exact. 


CHAP.  XV.]  THE   STONE   AECH.  311 


CHAPTEE   XY. 

THE    STONE   ARCH. 

167.  Definitions,  etc. — In  the  stone  arch  we  have  a  system 
of  bodies  in  contact  with  each  other,  and  so  supported  between 
certain  fixed  points,  that  they  are  not  only  in  equilibrium 
among  themselves,  but  also  with  the  exterior  forces.  The  sur- 
faces of  contact  we  call  the  bed-joints  ;  the  fixed  points  are 
the  abutments  ;  the  central  or  highest  arch  stone  is  called  the 
key -stone,  and  those  resting  upon  the  abutments,  the  imposts. 
The  inner  and  outer  limiting  surfaces  of  the  arch,  generally 
curved,  are  designated  as  the  intrados  and  extrados,  and  the 
arch  stones  generally  are  called  voussoirs. 

16§.  L.IIIC  of  Pre§sure§  in  Arch. — We  have  already  indi- 
cated (Art.  28,  Fig.  16)  the  manner  in  which  a  number  of  suc- 
cessive forces  are  resisted  by  an  arch.  We  see  from  the  force 
polygon  in  that  Fig.  that  the  horizontal  pressure  is  the  same  at 
every  point,  and  that  the  vertical  pressure  is  equal  to  the  sum 
of  the  weights  between  the  crown  and  any  point.  The  pres- 
sure line  is  then  an  equilibrium  polygon  formed  by  laying  off 
the  weights  of  the  arch  stones,  choosing  a  pole,  and  drawing 
lines  from  this  pole,  etc.,  as  described  in  our  second  chapter. 

If  the  weights  are  very  small,  and  their  number  very  great, 
the  equilibrium  polygon  becomes  a  curve.  This  curve  for 
equilibrium  should  never  pass  outside  the  limits  of  the  arch. 

169.  Sliding  of  the  Arch  Johns. — The  arch  is  properly, 
then,  nothing  but  a  curved  wall.  Upon  a  vertical  wall,  which 
may  also  support  loads,  but  which  has  no  horizontal  thrust,  only 
vertical  forces  act,  and  the  resultant  is  known  in  position  and 
direction.  We  may,  then,  investigate  the  stability  of  an  ordi- 
nary wall,  and  apply  the  results  directly  to  the  arch. 

We  assume  the  wall  divided  by  plane  bed-joints  extending 
through  its  entire  breadth,  whose  distances  apart  depend  upon 
the  dimensions  of  the  stones.  These  joints  are  the  weak  places 
of  the  wall,  since  separation  here  is  not  resisted  by  the  greatest 
strength  of  the  stone.  Neglecting  the  influence  of  the  mortar, 


312  THE    STONE   ARCH.  [CHAP.  XV. 

we  assume  that  any  section  along  a  bed-joint  resists  only  a  per- 
pendicular pressure  due  to  the  parts  above,  and  a  force  paral- 
lel to  the  joint  which  must  not  exceed  the  resistance  to  sliding 
due  to  friction.  If  this  parallel  force  is  greater  than  the  resist- 
ance of  friction,  the  upper  part  will  slide  upon  the  joint. 

If  we  represent  the  greatest  angle  of  repose  by  <£,  then  the 
resultant  of  the  vertical  forces,  acting  upon  the  joint  in  ques- 
tion, must  make  an  angle  with  the  normal  to  the  joint  less  than 
the  angle  <£.  Thus  at  the  joint  A  (Ph  24,  Fig.  95),  this  angle  is 
greater  than  </>,  and  the  upper  part  will  slide  along  this  joint. 
At  B  this  angle  is  less  than  </>,  and  no  sliding  can  take  place. 

The  ratio  of  the  force  of  friction  due  to  the  component  of  P 
normal  to  the  joint,  to  the  component  of  P  parallel  to  the  joint, 
we  call  the  coefficient  of  safety  against  sliding.  It  is  evidently 

equal  to — ,  or  to  the  distance  G-N  divided  by  PN. 

tan  P  N 

Since  we  can  dispose  the  bed-joints  at  pleasure,  we  may, 
always  make  them  perpendicular  to  the  direction  of  the  pres- 
sure, for  instance  in  Fig.  95  horizontal ;  or  at  least  so  place 
them  that  their  normals  vary  from  the  direction  of  the  resultant 
of  the  outer  forces,  at  most  by  an  allowable  angle  P  N. 

The  sliding  of  the  joints  can  then  always  be  prevented  by  the 
position  of  the  bed-joints. 

17O.  Force§  acting  upon  a  Cros§-§ection— Neutral  Axis. 
— Let  us  consider  what  happens  when  the  resultant  of  the  outer 
forces  acting  upon  a  joint,  instead  of  acting  at  the  centre  of 
gravity,  approaches  the  edge  of  a  joint,  under  the  assumption 
that  sliding  cannot  take  place,  or  that  the  direction  of  this  re- 
sultant is  perpendicular  to  the  joint.  There  is  no  reason  for 
assuming  the  distribution  of  pressure  upon  the  joint  surface 
any  different  from  the  case  of  a  beam.  The  stone,  as  well  as 
the  mortar,  is  elastic,  though  in  a  less  degree  than  wood  or  iron, 
and  accordingly  the  pressure  at  any  portion  of  the  joint  is  pro- 
portional to  the  approach  of  the  limiting  surfaces  of  the  upper 
and  lower  portions  of  the  wall.  If,  then,  we  assume  that  these 
surfaces  are  plane  before  and  after  loading,  if  the  resultant 
pressure  does  not  act  at  the  centre  of  gravity,  but  near  to  one 
edge,  the  pressure  at  different  points  will  vary,  and  there  will 
be  a  neutral  axis,  or  line  of  no  pressure,  either  within  or  wholly 
t  without  the  joint  surface. 

Every  cross-section  is  therefore  acted  upon  by  a  system  of 


CHAP.  XV.]  THE    STONE   ARCH.  313 

parallel  forces  whose  intensities  are  directly  as  their  distances 
from  a  certain  axis. 

Now,  neglecting  the  influence  of  the  mortar,  the  wall  can 
resist  compression  only.  No  tension  can  exist  at  any  point  of 
the  joint  surface. 

Clearly,  then,  the  neutral  axis  should  lie  wholly  without  the 
cross-section,  or  at  most  only  touch  it.  It  should  never  be 
found  within  the  cross-section,  as  in  that  case  all  the  material 
on  the  other  side  is  useless,  and  might  be  removed  entirely 
without  affecting  the  pressure  upon  the  actual  bearing  surface. 

The  neutral  axis,  then,  should  always  lie  without  the  cross- 
section  of  the  joint. 

IT  I.  System  of  Parallel  Forces  wlu»se  Intensities  are 
proportional  to  their  Distances  from  a  certain  Axis— The 
Kernel  of  a  Cross-section. — If  in  a  system  of  equal  and  paral- 
lel forces  we  find  the  moment  of  each  of  these  forces  with 
reference  to  a  certain  axis,  and  then  consider  these  moments  as 
themselves  forces,  we  shall  have  a  system  of  the  kind  referred 
to,  since  each  moment  force  will  be  directly  proportional  to  its 
distance  from  a  given  axis. 

Now,  as  we  have  seen  in  Art.  60,  Chapter  VI.,  the  centre  of 
action  of  such  a  system  of  moment  forces  does  not  coincide 
with  the  centre  of  gravity  of  the  original  simple  forces,  but  for 
any  given  axis  is  found  from  the  central  curve  of  the  cross-sec- 
tion. In  PL  11,  Fig.  35,  we  have  already  given  the  construc- 
tion for  finding;  this  centre  of  action,  the  semi-diameter  of  the 

O  ' 

central  curve  being  known,  for  any  given  axis. 

Suppose  now  this  axis  to  envelop  in  all  its  different  posi- 
tions the  outline  of  the  given  cross-section,  and  find  the  corre- 
sponding positions  of  the  centre  of  action  of  the  moment  forces. 
These  different  points  lie  in  a  closed  figure  which  we  may  call 
the  kernel  of  the  cross-section.  Then,  in  order  that  we  may 
always  have  compression  in  every  part  of  the  joint  surface  of 
our  wall,  the  resultant  of  the  forces  acting  upon  it  should 
always  act  within  the  kernel. 

In  Plates  11  and  12,  Figs.  36,  37,  38  and  40,  we  have  con- 
structed the  kernels  of  the  various  cross-sections  represented. 

Thus  in  Fig.  36,  according  to  the  construction  of  Art.  62,  for 
an  axis  at  A,  we  describe  upon  O  C  a  semi-circle.  Then  with  O 
as  a  centre  and  radius  equal  to  semi-diameter  of  the  central 
ellipse  on  A  C,  describe  an  arc  intersecting  the  semi-circle  in  a. 


314  THE    STONE   AKCH.  [dlAP.  XV. 

From  a  drop  a  perpendicular  upon  A  C,  and  we  obtain  the 
centre  of  action  for  axis  at  A.  A  similar  construction  for  other 
axes,  as  A  B,  B  C,  etc.,  give  us  other  points,  and  we  thus  find 
the  small  central  parallelogram,  which  is  the  kernel  or  locus 
of  the  centres  of  action  of  the  moment  forces  for  all  positions 
of  the  axis  enveloping  the  parallelogram  A  B,  C  D.  A  similar 
construction  gives  us  the  kernel  for  the  other  figures. 
We  have  from  Art.  60 

a' 

m  =  —  > 

^ 

where  m  =  the  distance  of  the  resultant  P  of  the  forces  acting 
upon  the  cross-section  from  the  centre  of  gravity,  and  a  =  the 
semi-diameter  of  the  central  curve,  and  i  =  the  distance  of  the 
neutral  axis  from  the  parallel  diameter  of  the  central  curve. 

If  we  call  c  the  distance  of  an  outer  fibre  from  this  diameter 
measured  on  the  side  of  P,  its  distance  from  the  neutral  axis 
is  i  +  c.  If  the  strain  in  this  fibre  is  S,  we  have 

P 

s  -i  +  c  ;  ;  -r-  :  i, 

-A. 

where  A  is  the  area  of  the  cross-section.     Hence 


If  P  acts  at  the  centre  of  gravity  of  the  cross-section,  i=  oo 

p 

(Art.  60),  the  neutral  axis  is  infinitely  distant,  and  S  =  —  .     If 

A. 

P  moves  away  from  the  centre  of  gravity,  the  neutral  axis 
approaches,  and  is  always  parallel  to  the  conjugate  diameter 
in  the  central  ellipse.  When  P  reaches  the  perimeter  of  the 
kernel,  the  neutral  axis  touches  the  perimeter  of  the  cross- 
section,  and  at  least,  then,  in  one  point  of  this  perimeter,  the 
pressure  is  zero.  If  P  passes  beyond  the  kernel,  the  neutral 
axis  enters  the  cross-section,  and  tensile  strains  enter  on  one 
side  to  balance  the  compressive  strains  on  the  other.  The  ker- 
nel then  forms  a  limit  beyond  which  the  resultant  P  must  'not 
act. 

172.     Position  of  Kernel  for  different  Cross-sections.— 
If  the  cross-section  is  symmetrical  with  reference  to  the  cen- 

c  P 

tre  of  gravity,  we  have  —  .  =  1,  and  therefore  S  =  2  —  ;  that  is, 

1  A. 


CHAP.  XV.]  THE    STONE    AKCH.  315 

when  the  neutral  axis  touches  the  cross-section,  or  P  acts  in 
the  kernel,  the  strain  S  is  twice  as  great  as  when  P  passes 
through  the  centre  of  gravity  of  the  joint  surface  and  is  uni- 
formly distributed. 

As  P  passes  beyond  the  kernel,  the  neutral  axis,  as  we  have 
seen,  enters  the  joint  area,  and  on  the  side  away  from  P  occa- 
sions, or  would  occasion  in  a  beam,  tensile  strains.  But  as  the 
assumption  is  that  the  joint  (neglecting  mortar)  cannot  resist 
tensile  strains,  we  may  remove  all  that  portion  on  the  opposite 
side  of  the  neutral  axis  without  increasing  the  pressure  on  the 
other  side. 

In  this  case,  then,  the  central  ellipse  is  not  that  for  the  whole 
joint  area,  but  only  for  that  portion  011  the  same  side  as  P,  and 
P  is  upon  the  kernel  for  that  portion. 

This  portion  can  be  determined  directly  for  a  certain  posi- 
tion of  P  only  in  a  few  individual  cases  ;  generally,  it  must  be 
found  by  trial.  We  must  first  find  'for  the  central  ellipse  of 
the  entire  joint  area  the  neutral  axis  corresponding  to  given 
position  of  P,  and  then  draw  a  parallel  cutting  off  somewhat 
more  of  the  area.  Then  determine  the  central  ellipse  of  the 
cut-off  portion,  and  see  if  the  pole  lies  symmetrically  to  the 
pole  of  the  cutting  line. 

The  parallelogram  is  one  of  the  areas  in  which  we  can  de- 
termine directly  the  amount  cut  off  when  P  acts  at  a  point 
upon  the  line  joining  the  centres  of  two  opposite  sides.  For  if 
we  cut  off  by  a  parallel  to  these  sides  a  portion  so  that  P  is  at 
£d  of  the  line  joining  the  centres  of  the  opposite  sides  of  the 
new  parallelogram,  then  P  lies  upon  the  kernel  for  this  new 
area.  The  proof  is  easy.  The  moment  of  inertia  of  the 
parallelogram  is  -£%  b  A3,  with  reference  to  the  diameter  b.  The 
square  of  the  radius  of  gyration  aa  is  then  y1^  A2.  The  distance 
of  the  point  of  application  of  P  from  one  of  the  sides  is 

a" 

^  -h  m  =  ^  +  —.    Hence 


The  half  height  of  the  kernel  is,  then,  -J-th  the  height  of  the 
parallelogram,  or  the  kernel  occupies  the  inner  third.  (See  Fig. 
36  ;  also  Woodbury  :  Theory  of  the  Arch,  p.  328,  Art.  3.) 

For  any  given  position  of  P,  then,  three  times  its  distance 


316  THE  STONE  ARCH.  [CHAP.  XV*. 

from  the  nearest  side  on  a  line  parallel  to  the  other  two,  gives 
the  position  of  the  fourth  side  of  the  parallelogram  for  which 
P  is  upon  the  kernel. 

173.  The  resultant  pressure  should  therefore  act 
within  the  middle  third  of  the  joint  area. — As  this  prin- 
ciple is  most  important,  and  the  demonstrations  of  Chapter  VI., 
upon  which  the  above  result  is  based,  may  appear  to  some  too 
purely  mathematical,  we  give  here  the  demonstration  of  the 
same  principle  as  given  by  Woodbury,  in  the  work  above 
cited. 

"  Suppose  the  pressure  to  be  nothing  at  the  in  trad  os  #,  and 
to  increase  uniformly  from  that  point  to  the  extrados  b  (PL 
24,  Fig.  96).  It  is  plain  that  the  pressure  at  any  point  along  a  ~b 
will  be  represented  by  the  brdinate  of  a  certain  triangle.  The 
whole  pressure  will  be  represented  by  the  surface  of  that  tri- 
angle ;  and  the  point  of  application  of  the  resultant  of  all  the 
pressures  will  be  at  c  opposite  the  centre  of  gravity  of  that 
triangle.  We  have  then  c  b  =  ^  a  ft.  Vice  versa,  if  the  point 
of  application  be  at  c,  c  I  —  %  a  5,  we  know  that  the  pressure 
is  nothing  at  a. 

"  If  the  point  of  application  be  at  c,  c  7>  being  less  than  -J  a  J, 
c  being  still  opposite  the  centre  of  gravity  of  the  triangle 
whose  ordinates  represent  the  pressure,  we  know  that  the  ver- 
tex of  that  triangle  and  point  of  no  pressure  are  at  e}be  =  3 
xbc. 

"  In  this  case,  the  joint  a  b  will  open  at  a  as  far  as  e ;  the 
adjacent  joints  will  also  open  until  we  come  to  one  where  the 
curve  of  pressure  passes  within  the  prescribed  limit. 

"  This  reasoning  is,  of  course,  applicable  to  all  the  joints ; 
and  we  readily  conclude  that  the  curves  of  pressure  should  lie 
entirely  between  two  other  curves  which  divide  the  joint  into 
three  equal  parts." 

Thus,  in  PL  24,  Fig.  97,  suppose  .the  resultant  P  of  the 
upper  part  of  the  wall  to  have  the  position  as  represented,  so 
that  it  intersects  the  joint  B  D  in  C  outside  of  the  middle  third 
of  the  cross-section.  The  entire  pressure  is  distributed  over 
3  C  B  =  A  B,  and  the  area  D  A  does  not  act  at  all.  Moreover, 
the  pressure  at  B  is  twice  as  great  as  when  P  passes  through 
the  centre  of  gravity  and  is  uniformly  distributed  over  A  B,  or 
is  f  ds  of  the  uniformly  distributed  pressure  of  'P  upon  C  B. 

Beyond  A  the  pressure  is  zero,  and  the  conditions  of  load 


CHAP.  XV.]  THE    STONE   AKCH.  317 

and  equilibrium  would  not  be  changed  if  the  stone  beyond  A 
were  removed. 

If  C  approaches  still  nearer  B,  so  that  the  pressure  is  distrib- 
uted upon  an  ever-decreasing  area,  the  resistance  of  the  mortar 
will  be  finally  overcome  ;  it  will  be  forced  out,  and  stone  will 
come  in  contact  with  stone,  and  there  will  be  rotation  about  the 
edge  at  B.  This  rotation  can  never  occur  if  the  pressure  P  is 
distributed  over  the  whole  joint  area.  If,  then,  we  consider 
rotation  to  commence  at  the  moment  when  P  is  no  longer  dis- 
tributed over  the  entire  area — when,  therefore,  the  neutral  axis 
just  enters  the  joint — then,  in  order  that  no  rotation  may  occur,  y 
P  must  pierce  the  joint  area  inside  the  kernel. 

174.  LJne  of  Pressures  in  tlie  Arch. — When  the  dimen- 
sions and  form  of  a  wall  are  given,  we  can  determine  directly 
the  resultant  P  of  the  outer  forces  acting  upon  a  joint,  and  then 
by  the  two  preceding  Arts,  can  determine  the  condition  of  sta- 
bility of  the  wall.  In  the  arch,  however,  we  cannot  determine 
P  directly  for  a  given  cross-section,  but  must  first  make  certain 
assumptions. 

In  the  first  place,  it  is  clear  that  an  arch  is  stable  when  it  is 
possible  in  two  joints  to  take  two  reactions  P1  and  P2  (PI.  24, 
Fig.  98)  such  that,  with  the  weight  of  the  intervening  portion 
of  the  arch  and  its  load,  the  resulting  line  of  pressure  shall  lie 
so  far  in  the  interior  of  the  arch  that  rotation  about  a  joint  edge 
cannot  take  place.  If  the  arch  is  so  feeble  and  the  resistance 
of  the  material  so  slight  that  only  one  such  assumption  of  P^  and 
P2  can  be  made,  and  only  one  such  pressure  line  drawn,  this  is 
plainly  the  true  pressure  line  for  stability,  and  by  it  Pt  and  P2, 
as  also  the  pressure  at  every  joint,  are  determined. 

If,  however,  the  arch  is  so  deep  and  the  resistance  of  the 
material  so  great  that  by  variation  of  P1  and  P2  several  such 
pressure  lines  may  be  drawn,  none  of  which  causes  rotation 
about  a  joint  edge,  which  of  all  these  possible  pressure  lines  is 
the  true  pressure  line  of  the  arch  ? 

We  assert :  That  is  the  true  pressure  line  which  approaches 
nearest  the  axis,  so  that  the  pressure  in  the  most  compressed 
joint  edge  is  a  minimum. 

If  we  assume  the  material  so  s}ft  that  the  pressure  line  ap- 
proaches the  axis  so  near  that  only  one  assumption  of  P,  and 
P,  is  possible,  then  this  would  evidently  be  the  true  pressure 
line.  If  now  the  material  hardens  without  altering  any  of  its 


318  THE    STONE    ARCH.  [CHAP.  XV. 

other  properties,  such  as  its  specific  weight  or  modulus  of  elas- 
ticity, then  the  position  of  the  pressure  line  is  not  altered.  As 
there  is  no  reason  for  supposing  the  pressure  line  different  in 
an  arch  built  of  hard  material  from  that  in  one  originally  soft 
which  has  afterwards  gradually  hardened,  it  follows  that  the 
pressure  line  in  all  arches  of  same  form  and  loading  has  the 
same  position  which  it  would  have  had  if  the  arch  had  been 
originally  of  the  softest  material ;  that  is,  that  position  which 
makes  the  pressure  in  the  most  compressed  joint  edge  a  mini- 
mum. 

In  order  to  draw  the  pressure  line  in  an  arch,  we  may  then 
seek  by  means  of  the  formula 

o  •* 

this  pressure  in  the  joint,  where  the  pressure  line  approaches 
nearest  the  edge,  and  ascertain  whether  it  can  be  still  further 
diminished  by  change  of  position  of  the  pressure  line.  This  is, 
however,  not  necessary.  We  have  only  to  ascertain  whether  it 
is  possible  to  draw  a  pressure  line  whose  sides  cut  the  corre- 
sponding joint  a.rea,  within  the  kernel,  for  then,  since  we  know 
that  there  can  be  a  still  more  favorable  position,  there  is  no 
danger  of  rotation. 

175.  The  Line  of  Support. — The  curve  formed  by  joining 
the  intersections  of  the  sides  of  the  pressure  line  with  the  joint 
areas  we  call  the  support  line,  or  line  of  support. 

If  the  joints  of  an  arch  answer  to  the  condition  of  Art.  169, 
so  that  sliding  of  the  joints  cannot  occur,  we  see  at  once  from 
the  position  of  the  support  line  on  what  side  and  where  rota- 
tion will  take  place.  If  at  any  point  this  line  passes  beyond 
the  kernel,  we  have  theoretical  beginning  of  rotation ;  if  it 
passes  outside  of  the  arch,  there  is  actual  rotation,  and  if  it  lies 
within  the  kernel,  there  is  no  rotation. 

The  manner  of  determining  from  the  position  of  the  support 
line  all  the  possible  motions  of  an  arch  is  illustrated  in  the  fol- 
lowing Figs. 

In  PI.  24:,  Fig.  99,  we  have  a  possible  support  line  touching 
the  extrados  at  crown  and  springing,  and  the  intrados  between 
these  points.  We  have  accordingly  rotation  at  crown,  and  at 
the  points  between  crown  and  springing,  so  that  the  joints  at 
these  points  open  on  the  sides  of  the  arcli  opposite  the  support 
line.  The  crown  will  sink,  and  as  at  the  crown  and  flanks  the 


CHAP.  XV.]  THE  STONE  AECH.  319 

support  line  is  approximately  parallel  to  the  extrados  and  intra- 
dos,  there  will  be  several  joints  in  the  same  condition,  and 
several  will  open,  as  indicated  in  the  Fig. 

In  PI.  25,  Fig.  100,  we  have  the  condition  of  stability  of  a 
pointed  arch,  not  loaded  at  the  crown.  The  support  line  ia 
horizontal  at  crown,  and  there  is  no  angle  there,  as  in  the  arch 
itself.  The  rotation  at  various  points  is  indicated  in  the  Fig. 
We  shall  soon  see  that  the  support  line  deviates  but  very  little 
from  the  pressure  line.  From  the  direction  of  the  tangent  to 
the  support  line  at  any  point,  therefore,  we  may  conclude  as  to 
the  conditions  of  sliding. 

From  Fig.  101  we  may  conclude  that  the  arch  will  slide  out- 
wards upon  the  right  abutment.  The  rotation  at  various  points 
is  given  by  the  Fig.  It  is  sufficient,  as  we  see,  to  make  the 
abutment  surface  more  nearly  perpendicular  to  the  support 
line,  as  shown  in  the  left  abutment,  to  prevent  this  sliding,  and 
at  the  same  time  a  more  favorable  support  line  can  be  drawn. 
Since,  as  we  have  seen  in  Art.  100,  sliding  can  and  must  in 
this  manner  be  always  prevented,  we  shall  give  no  more  exam- 
ples of  arches  unstable  in  this  particular. 

The  arches  of  Figs.  99  and  100  can  be  made  stable  by  suffi- 
ciently increasing  their  thickness,  or  conforming  their  shape 
more  nearly  to  that  of  the  support  line. 

176.  Deviation  of  the  Support  from  the  Pre§sure  Line. 
This  deviation  is  not  great.  In  order  to  make  it  apparent,  we 
must  draw  a  pressure  line  for  slight  pressure  in  the  lower  part 
of  an  arch  with  very  long  and  inclined  voussoirs  [PL  25,  Fig. 
102].  Thus,  if  we  combine  the  weights  of  the  voussoirs  1, 2,  3, 
4,  etc.,  acting  at  their  centres  of  gravity,  with  the  pressure^Q  in 
the  first  joint,  we  have  the  pressure  line  shown  by  the  broken 
line  1,  2,  3,  4,  5,  6,  7,  8,  whose  sides  1  2,  2  3,  3  4,  etc.,  give  the 
direction  of  the  pressure  in  the  corresponding  joints  between 
the  voussoirs  1  and  2,  2  and  3,  etc.  Thus  5  6  is  the  direction 
of  the  pressure  upon  the  joint  between  voussoirs  5  and  6.  This 
direction  cuts  the  joint  at  5',  which  is  therefore  the  point  of 
application  of  the  pressure,  or  a  point  upon  the  line  of  support. 
Thus  we  find  3',  4',  5',  6',  and  the  line  joining  these  points  is 
the  support  Iww.  In  general,  then,  the  support  and  pressure 
lines  coincide  when  the  vertical  through  the  centre  of  gravity 
of  any  very  small  element  coincides  with  the  joint,  and  they 
deviate  when  this  vertical  does  not  coincide  with  the  joint. 


320  THE    STONE    ARCH.  [CTIAP.  XV. 

In  the  ordinary  form  of  joint,  as  shown  in  Fig.  103,  the  sup- 
port line  varies  from  the  pressure  line,  since  the  vertical  through 
the  centre  of  gravity  S  does  not  coincide  with  the  joint  under 
S.  If,  however,  we  should  conceive  the  arch  divided  into  ver- 
tical laminae,  then  the  support  and  pressure  lines  fall  together. 
This  is  precisely  the  assumption  always  made  in  the  analytical 
discussion  of  the  theory  of  the  arch. 

Thus  we  take  the  area  A  =  /  y  dx,  and  this  expression  sup- 
poses the  arch  divided  into  vertical  laminse. 

The  first  to  make  clearly  this  distinction  between  the  lines  of 
pressure  and  support,  was  Mosely  (Civil  Eng.).  Other  authors 
have  after  him  adopted  this  distinction,  and  then  proved  that 
the  two  lines  always  coincide,  without  remarking  that  this  coin- 
cidence is  only  because  of  the  adoption  of  the  above  integral. 
The  same  assumption  simplifies  greatly  the  graphical  construc- 
tion also  (the  analytical  treatment  is  without  it  well-nigh  im- 
possible). We  shall  therefore  assume  vertical  lam i  1133  where 
it  is  at  all  permissible.  This  is  always  permissible  at  the  crown 
of  arches  with  horizontal  tangent,  because  there  the  joints  are 
vertical,  and  over  all,  when  the  pressure  line  lies  below  the 
axis  of  the  arch ;  for  the  support  line  lies  always  above  the 
pressure  line,  and  therefore,  in  this  case,  the  conditions  of  sta- 
bility are  more  favorable  for  it  than  for  the  pressure  line  itself, 
when  considered  as  the  line  of  support. 

Moreover,  it  is  easy  at  any  point  of  the  pressure  line  con- 
structed with  vertical  laminae  to  pass  to  that  line  for  another 
form  of  joint,  and  to  the  corresponding  support  line.  Thus,  if 
for  the  point  A  (PL  25,  Fig.  104)  we  have  found  the  pressure 
Q,  and  if  now  we  wish  to  pass  to  the  joint  ABC,  we  prolong 
Q  till  it  meets  P,  the  weight  of  the  voussoir  A  B  C  D,  and  re- 
solve P  and  Q  at  this  point  into  Q'.  Then  Q'  is  a  side  of  the 
new  pressure  line,  and  it  cuts  A  B  in  a  point  of  the  support 
line. 

In  this  way  we  can  easily  determine  whether  the  error  com- 
mitted when  we  substitute  the  pressure  line  for  vertical  laminae 
for  that  for  the  actual  joints,  which  is  given  by  the  segment  of 
the  joint  A  B  between  QJ.  and  the  pressure  line,  can  be  disre- 
garded. 

177.  Dimension*  of  the  Arcli. — The  object  of  the  con- 
struction of  the  pressure  or  support  line  in  the  arch  is  to  deter- 


CHAP.  XV.]  THE    STONE    ARCH.  321 

mine  the  stability  and  the  joints  of  the  abutments.  When,  the 
live  load  of  the  arch  can  be  neglected  with  respect  to  its  own 
weight,  and  when  the  material  of  the  arch  possesses  the  usual 
strength,  and  the  pressure  line  lies  within  the  inner  third,  then 
the  lower  point  of  rupture  lies  so  low  that  the  back  masonry 
reaching  from  this  point  beyond  the  pressure  line  completely 
encloses  it. 

There  is,  therefore,  nothing  arbitrary,  when  the  form  of  the 
arch  is  given,  except  the  depth.  Since  in  an  arch  of  less  depth 
than  is  allowable  in  practice  a  support  line  can  still  be  in- 
scribed, the  graphical  method  is  unable  to  determine  the 
proper  depth.  We  must  then  leave  to  theory  the  development 
of  formulae  by  which  this  can  be  determined,  and  assume  that 
not  only  the  form  of  the  arch  is  given,  but  also  its  proper 
depth  and  the  lower  joint  of  rupture.  It  is  required  to  deter- 
mine the  stability  of  the  abutments. 

The  stability  of  the  abutments  can  be  regarded  from  two 
points  of  view.  We  may  consider  it  as  a  continu|tion  of  the 
arch,  as  in  many  English  and  French  bridges,  in  which  the 
arch  is  continued  as  such,  clear  to  the  foundation  ;  or  we  may 
regard  it  as  a  wall  whose  moment  about  the  joint  of  rupture 
resists  the  rotation  about  this  joint  due  to  the  thrust.  Both 
views  are  identical,  as  the  entire  theory  of  the  support  line  rests 
upon  the  investigation  of  the  rotation.  They  differ  only  in  the 
method  of  expressing  the  safety  of  the  abutment. 

If  the  arch  is  continued  to  the  foundation,  and  the  space  be- 
tween it  and  the  road  line  filled  up  by  spandrels ;  or  if  the 
thickness  of  the  abutment  increases  from  above  as  the  support 
line  requires  ;  or,  as  is  often  the  case  in  England,  the  abutment 
consists  of  walls  parallel  to  the  crown,  separated  by  hollow 
spaces;  still,  in  every  case  the  abutment  is  not  to  be  distin- 
guished from  the  arch  proper — it  is  stable  when  the  support 
line  lies  in  the  interior.  If  the  prolonged  arch  is  separated 
entirely  from  the  adjacent  masonry,  there  is  no  reason  for  not 
giving  the  axis  of  the  prolongation  the  form  of  the  support 
line  itself. 

If,  on  the  other  hand,  there  is  no  separation  of  the  arch  and 
abutment,  as  in  the  English  hollow  abutments,  it  is  sufficient 
that  the  support  line  lie  in  the  inner  third,  and  the  abutment 
will  be  certainly  stable. 

The  supposition  that  the  resistance  of  the  rnortar  is  suffi- 
21 


322  THE    STONE   ARCH.  [CTIAP.  XV. 

cientlj  great  to  unite  the  whole  abutment  as  a  single  block 
which  turns  about  its  under  edge,  gives  too  small  dimensions. 
To  ensure  safety  it  is  assumed  that  equilibrium  exists  with 
reference  to  rotation  about  the  lower  edge,  when  the  thrust  of 
the  arch  is  1.5  greater  than  the  actual.  Investigations  of 
French  engineers  have  shown  that  this  coefficient  of  safety  for 
very  light  arches  is  not  less  than  1.4.  The  old  tables  of  Petit 
give  1.9.  We  assume  it,  therefore,  =  2. 

If,  therefore,  the  double  thrust  of  the  arch  at  the  lower  point 
of  rupture  is  united  with  the  weight  of  the  abutment,  the  re- 
sultant should  still  fall  within  the  base.  Since  it  is  indiiferent 
in  what  order  the  elements  of  the  abutment  are  resolved,  it  is 
best  to  divide  it  into  vertical  laminae,  and  unite  these  with  the 
double  thrust.  The  equilibrium  polygon  thus  obtained  should 
cut  the  foundation  base  within  the  edge  of  the  abutment. 

When  the  thickness  of  the  abutment  is  thus  determined,  we 
must  construct  the  actual  pressure  line  for  the  simple  thrust 
in  order  to  Determine  the  joints.  In  drawing  this  second  pres- 
sure line,  we  should  properly  take  the  divisions  of  the  arch  by 
the  joints  themselves.  If,  however,  we  take  the  division  in 
vertical  laminae,  the  deviation,  as  we  have  seen,  is  insignificant. 
The  normals  to  the  actual  joints  must,  then,  not  deviate  from 
the  sides  of  this  pressure  line  by  more  than  the  angle  of 
repose. 

178.  Construction  of  the  Pressure  Line. — In  PL  25,  Fig. 
105,  we  give  the  method  of  construction  of  the  proper  width  of 
abutment  for  an  arch.  We  first  divide  the  arch  into  vertical 
laminse,  and  determine  their  weight.  If  the  surcharge  has 
vacant  spaces,  or  is  generally  of  different  specific  weight  from 
the  material  of  the  arch  itself,  it  must  first  be  reduced.  Thus, 
if  the  surcharge  (spandrel  filling,  etc.)  weighs,  for  instance,  only 
f  ds  as  much  as  an  equal  area  of  masonry  in  the  arch,  we  have 
simply  to  diminish  the  vertical  height  above  the  arch  by  -Jd. 
We  thus  obtain  the  dotted  line  given  in  the  Fig.,  which  forms 
the  limit  of  the  reduced  laminae,  and  we  can  treat  the  areas 
bounded  by  this  line — the  vertical  lines  of  division  and  the 
intrados — as  homogeneous.  We  have  then  only  to  determine 
the  centres  of  gravity  of  the  various  laminae  according  to  the 
construction  for  finding  the  centre  of  gravity  of  a  trapezoid 
(Art.  33),  and  suppose  at  these  points  the  weights,  which  are 
proportional  to  the  reduced  areas  of  the  trapezoids  to  act. 


CHAP.  XV.]  THE    STONE   ARCH.  .    323 

Laying  off  these  weights  in  their  order,  we  have  the  force 
line  (Fig.  to  left).  The  weights  of  the  abutment  laminae  9,  10 
and  11  are  laid  off  to  same  scale  one-half  'of  their  proper  in- 
tensities. The  reason  will  soon  appear. 

1st.  To  determine  the  thrust  H,  and  also  the  joint  of  rup- 
ture. 

We  first  inscribe  a  pressure  line  by  eye,  and  assume  the 
point  of  the  intrados  to  which  this  line  most  nearly  approaches 
as  the  edge  of  the  joint  of  rupture.  Draw  next  from  the  cor- 
responding point  of  the  force  line  a  line  parallel  to  the  assumed 
pressure  line  at  this  point.  This  line  will  cut  off  from  the 
horizontal  through  the  beginning  of  the  force  line  our  first 
approximate  value  of  H. 

Thus,  suppose  we  have  inscribed  by  eye  the  pressure  line  1, 

2,  3,  4,  etc.,  which  gives  us  the  point  a  for  the  position  of  the 
edge  of  the  joint  of  rupture.     Then  a  line  drawn  from  5  on 
the  force  line,  parallel  to  the  side  4  5  of  the  pressure  line,  gives 
us  our  first  value  for  H. 

Now  assuming  this  value  of  H,  we  erase  the  first  assumed 
pressure  line,  and  proceed  to  construct  the  pressure  line  cor- 
responding to  this  value  of  H,  and  the  force  line  divisions  1,  2, 

3,  4,  etc.     If  this  pressure  line  lies  always  within  the  middle 
third  of  the  arch,  it  may  be  taken  as  the  proper  pressure  line, 
and  H  as  the  true  thrust.     In  general,  however,  this  will  not 
be  the  case.     The  pressure  line  thus  determined  may  even  pass 
without  the  arch  entirely.     We  then  determine  the  new  point 
of  rupture,  as  given  by  the  point  of  exit  of  this  pressure  line, 
and  produce  the  side  at  this  point  back  to  intersection  with  H 
prolonged.     From  this  point  of  intersection  draw  a  line  which 
does  lie  within  the  middle  third  of  the  arch  at  the  lamina  of 
rupture,  and  then  in  the  force  polygon  from  the  corresponding 
point  of  the  force  line  draw  a  parallel  to  this  line,  thus  cutting 
off  a  new  value  for  H.     Erasing  now  the  preceding  pressure 
line,  we  construct  a  third  with  this  new  value  of  H,  which  will 
in  general  give  us  a  pressure  line  lying  everywhere  within  the 
middle  third  of  the  arch.     If  not,  another  approximation  may 
be  made.     We  thus  find  by  successive  approximation  the  true 
joint  of  rupture  and  the  corresponding  thrust. 

2d.  Width  of  abutment. — Since  we  have  laid  off  the  arch 
weights  to  scale  in  their  true  value,  the  pressure  line  thus  ob- 
tained is  the  true  pressure  line  for  the  arch.  But  we  have  laid 


324  THE    STONE    ARCH.  [CHAP.  XV. 

off  the  abutment  laminse  9,  10  and  11,  one-half  \\\G\Y  true  value, 
and  the  pressure  line  thus  obtained  with  the  same  thrust  and 
pole  O  is  the  same  as  if  we  had  taken  their  true  value  and  twice 
H.  Its  intersection  with  the  foundation  gives  us,  then,  the 
proper  width  of  the  abutment  for  stability,  according  to  our 
assumption  of  2  for  the  coefficient  of  stability  (Art.  177). 

179.  Thus  we  can  easily  determine  for  any  given  case  of 
arch  and  surcharge  the  horizontal  thrust  and  the  proper  width 
of  abutment,  and  then  from  the  pressure  line  can  easily  so  dis- 
pose the  joints  as  to  prevent  sliding.  If  the  dimensions  of  the 
arch  as  given  are  not  such  as  to  be  stable,  it  will  be  found  im- 
possible to  inscribe,  as  above,  a  pressure  line  which  shall  lie 
within  the  middle  third,  and  the  curve  of  'extrados  or  intrados 
will  have  to  be  altered  so  that  this  shall  be  the  case.  The  pres- 
sure line  thus  obtained,  it  is  true,  does  not  exactly  correspond 
with  the  true  one,  as  it  is  still  possible  to  inscribe  another  which 
shall  deviate  still  less  from  the  true  line.  We  have  also  taken 
the  double  thrust  for  the  abutment  laminse  alone,  instead  of 
for  all  laminse  from  the  joint  of  rupture  of  the  arch.  Both 
deviations  are  made  on  account  of  the  far  greater  ease  and 
rapidity  of  construction.  It  would  be  found  very  tedious  to 
take  first  the  force  polygon  up  to  somewhere  about  the  section 
of  rupture,  then  by  long  trial  find  the  innermost  support  line, 
and  finally,  after  the  section  of  rupture  is  by  this  line  deter- 
mined, to  lay  off  the  remainder  of  the  force  polygon,  and  pro- 
long the  pressure  line  through  the  abutments. 

It  is  far  simpler  to  'proceed,  as  above,  by  assuming  the  point 
of  application  of  the  horizontal  thrust,  as  also  temporarily  the 
section  of  rupture.  We  obtain  thus  a  somewhat  smaller  value 
for  the  width  of  abutment,  but,  on  the  other  hand,  we  have 
taken  the  coefficient  of  stability  at  2  instead  of  1.9,  as  assumed 
in  Petitfs  tables. 

Moreover,  the  widths  of  abutment  thus  obtained  are  greater 
than  those  obtained  by  these  tables,  as  it  is  assumed  in  them 
that  the  point  of  application  of  the  horizontal  thrust  is  at  the 
upper  edge  of  the  abutment.  Thus  in  every  respect  the  con- 
struction gives  results  reliable  and  even  more  accurate  than  the 
tables,  as  we  take  the  arch  as  it  really  is  in  any  given  case, 
while  in  the  tables  suppositions  are  made  with  reference  to 
spandrel  filling,  etc.,  which  do  not  hold  good  for  every  case. 

1§O.  Proper  Thickness  of  Arch  at  Crown. — The  proper 


CHAP.  XV.]  THE    STONE    AKCH.  325 

depth  of  the  arch  at  the  key  depends  not  only  upon  the  rise 
and  span,  but  also  upon  the  load.  The  pressure  at  the  extrados 
at  the  key,  which,  is  in  general,  the  most  exposed  part  of  joint, 
should  not,  according  to  the  best  authorities,  exceed  -^th  the 
ultimate  resisting  power  of  the  material.  If  P  is  the  pressure 
per  unit  of  surface,  H  the  thrust,  and  d  the  depth  of  key-stone 

2H 

joint,  then  P  —  -y, 

since,  on  the  assumption  that  the  curve  of  pressure  does  not  pass 
outside  the  kernel,  the  maximum  pressure  is  twice  the  mean 

TT 

pressure  -j.     This  mean  pressure,  then,  should  not  exceed  ^th 

CL 

the  ultimate  resistance  of  the  material.  In  the  best  works  of 
Rennie  and  Stevenson  the  thickness  at  key  varies  from  -^-th  to 
•gJg-d  the  span,  and  from  ^-th  to  ^th  tne  radius  of  the  intrados. 
The  augmentation  of  thickness  at  the  springing  line  is  made  by 
the  Stevenson's  from  20  to  40  per  cent.,  by  the  Rennie's  at 
about  100  per  cent. 

Perronet  gives  for  the  depth  at  crown  the  empirical  formula 

d=  0.0694 /*  +  0.325  meters, 

in  which  r  is  the  greatest  length  in  meters  of  the  radius  of  cur- 
vature of  the  intrados. 

For  arches  with  radius  exceeding  15  meters,  this  gives  too 
great  a  thickness.  According  to  Rankine, 

d  =  0.346  Vr 
for  circular  arches,  and 

d  =  0.412  Vr, 

where  r  is  the  radius  of  curvature  of  the  intrados  at  the  crown 
in  feet. 

"  The  London  Bridge  is  in  its  plan  and  workmanship  per- 
haps the  most  perfect  work  of  its  kind.  The  intrados  is  an 
ellipse,  the  span  152  ft.,  the  rise  £th  as  much,  the  depth  of  key 
-gig-th  the  span.  The  crown  settled  only  two  inches  upon  remov- 
al of  centres." — [  Woodbury  :  Theory  of  the  ArchJ] 

In  general,  we  must  first  assume  the  depth  at  key  in  view  of 
the  strength  of  the  material,  the  character  of  the  workmanship, 
the  load,  etc.  Then  the  thrust  being  found,  we  find  the  mean 
pressure  per  unit  of  area  as  above.  If  this  mean  pressure 
exceeds  ^th  the  ultimate  resisting  power  of  the  material,  make 


326  THE    STONE    ARCH.  [CHAP.  XV. 

a  new  supposition,  increase  the  thickness,  find  the  thrust  and 
pressure  anew,  and  so  on,  till  the  results  are  satisfactory. 

The  ultimate  resisting  power  of  granite  may  be  taken  at 
6,000  Ibs.,  brick  1,200,  sandstone  4,000,  limestone  5,000  Ibs. 
per  square  foot.  These  values  are,  of  course,  very  general, 
and  subject  to  considerable  variations,  according  to  the  kind 
and  quality  of  the  stone.  The  strength  of  the  material  to  be 
used  must,  for  any  particular  case,  be  determined  by  actual 
experiment. 

The  weight  of  a  cubic  foot  of  stone  may,  in  general,  be  as- 
sumed at  160  Ibs.,  brick  masonry  at  125  Ibs. 

IN!.  Increase  of  thickness  due  to  change  of  form.— 
Having  obtained  a  thickness  which  satisfies  all  the  conditions, 
we  must,  if  the  arch  be  very  light,  make  some  further  provi- 
sion for  the  change  of  form  which  is  sure  to  take  place  af ter 
the  removal  of  the  centres.  By  this  change  of  form  the  pres- 
sure line  is  altered,  and  the  thickness  may  need  to  be  increased. 
In  general,  we  need  only  to  increase  the  depth  from  the  key  to 
the  springing.  This  increase  need  not  exceed  fifty  per  cent,  at 
the  joint  of  rupture  and  weakest  intermediate  joint.  [  Wood- 
l>ury  :  Theory  of  the  Arch.] 

182.  Thus,  by  a  simple  and  rapid  construction,  we  can  de- 
termine, for  any  particular  case,  the  thrust,  joint  of  rupture, 
and  proper  thickness  of  the  abutments,  without  the  use  of 
tables  or  the  intricate  formulae  usually  employed.  There  is  no 
difficulty  in  laying  down  on  paper  and  verifying  all  the  ele- 
ments of  the  most  complex  case.  The  method  is  entirely  in- 
dependent of  all  particular  assumptions,  and  is  therefore 
especially  valuable  when  irregularities  of  outline  or  construc- 
tion place  the  arch  almost  beyond  the  reach  of  calculation.  It 
is  general,  and  may  be  applied  with  equal  ease  to  loaded  and 
unloaded,  full  circle,  segmental,  or  elliptical  arches  with  any 
form  of  surcharge. 


CHAP.  XVI.]  THE  INVESTED  AKCH.  327 


CHAPTEK   XYI. 

THE   INVERTED   AKCH SUSPENSION    SYSTEM. 

183.  The  inverted  arch  forms  the  supporting  member  of 
chain  or  cable  suspension  bridges.  Whether  the  cable  be  com- 
posed of  chains,  links,  or  wires,  we  suppose  them  so  flexible 
that  they  can  perfectly  assume  the  curve  of  equilibrium.  As, 
therefore,  disregarding  the  dead  weight,  any  partial  load  would 
cause  a  change  of  shape,  the  cables  must  be  stiffened  in  order 
to  prevent  the  motion  which  would  otherwise  take  place. 

We  may  stiffen  the  chains,  as  shown  in  PL  26,  Fig.  106,  by  trian- 
gular bracing,  thus  making  a  rigid  system  ;  or  we  may  have  two 
parallel  chains  and  brace  them  to  each  other,  as  shown  by  Fig. 
90  inverted ;  or  we  may  introduce  an  auxiliary  truss,  the  office 
of  which  is  not  to  add  in  any  degree  to  the  supporting  power 
of  the  combination,  but  simply  to  distribute  a  partial  load  over 
the  whole  span,  so  as  to  cause  it  to  take  effect  as  a  distributed 
load,  and  thus  prevent  change  of  shape. 

As  in  the  first  and  last  cases  the  structure  is  commonly 
hinged  at  the  centre  in  order  to  eliminate  the  effects  of  tem- 
perature, the  method  of  resolution  of  forces  explained  in  Arts. 
8-13  will,  in  general,  be  applicable  for  the  determination  of 
the  strains. 

In  the  second  case,  we  can  apply  the  principles  of  Arts.  158- 
161.  ! 

The  rear  chains,  anchorages,  and  stiffening  truss  deserve, 
however,  special  notice. 

1§4.  Rear  Chain*  and  Anchorages. — The  greatest  ten- 
sion in  the  main  chains  occurs,  of  course,  for  full  load.  To 
find  the  tension^at  top  of  tower,  as  also  the  horizontal  pull,  we 
have  simply  to  lay  off  half  the  whole  load  vertically  from  o  to 
d  [PI.  26,  Fig.  106],  and  then  draw  O  o  horizontal  and  O  d 
parallel  to  the  last  side  at  tower.  Then  O  d  is  the  tension  in 
that  side,  and  o  O  the  horizontal  pull.  This  pull  is  neutralized 
by  the  opposite  and  equal  pull  of  the  rear  chain  leading  to  the 
anchorage ;  provided,  as  should  always  be  the  case,  it  makes  an 


328  THE  INVERTED  ARCH.  [CHAP.  XVI. 

equal  angle  with  the  vertical.  We  have  thus  acting  upon  the 
tower  simply  the  half  load ;  and  the  tension  in  the  rear  chain 
is  equal  to  that  in  the  last  link,  O  d. 

If  from  O  we  draw  parallels  to  the  other  links,  we  have  at 
once  the  strains  in  these  links,  O  <?,  O  5,  O  a,  etc. 

Now,  if  the  anchorage  is  a  solid  block  of  masonry,  its  con- 
dition of  stability  is,  of  course,  very  easily  determined.  The 
moment  of  the  tension  in  the  rear  cable,  with  reference  to  the 
edge  of  rotation,  must  be  more  than  balanced  by  the  moment 
of  the  weight  of  the  block  acting  at  its  centre  of  gravity,  with 
reference  to  this  edge.  The  case  is  too  simple  to  need  further 
notice. 

It  is,  however,  more  economical  to  make  the  anchorage  hol- 
low— that  is,  in  the  form  of  an  arch.  The  preceding  method 
for  determining  the  stability  of  the  arch  has  then  here  direct 
application. 

Thus,  laying  off  along  the  vertical  through  the  centre  of  the 
tower  the  weights  of  segments  of  the  arch,  we  form  with  these 
segment  weights  and  the  double  tension  in  the  chain  an  equi- 
librium polygon.  For  this  we  have  the  pole  On  A  O1  being 
double  the  tension  Od  already  found.  We  then  draw  Otl,  (^2, 
Oj3,  etc.,  and  then  from  A  parallels  to  these  to  the  segment 
verticals  1,  2,  3,  etc.  We  thus  have  the  polygon  A  1,  2,  3,  4,  5. 
\_N~ote. — We  take  the  double  tension,  as  before,  for  the  arch,  we 
took  2  H  instead  of  H,  in  order  to  ensure  stability.] 

The  last  line  of  this  polygon  4  5  prolonged  must,  for  sta- 
bility, pass  within  the  pier  abutment,  and  its  resultant,  when  it 
is  combined  with  the  weight  of  the  pier  and  pier  abutment, 
must  pass  within  the  abutment  foundation.  Through  its  inter- 
section with  the  vertical  line  through  the  axis  of  the  tower  the 
curve  of  pressure  for  the  arch  must  pass. 

Drawing  now  Oa  4  parallel  to  the  rear  chain,  and  making  it 
also  equal  to  the  double  tension,  or  twice  O  d,  we  find  the  pole 
O2,  and  from  it  draw  O2 1,  O2  2,  O2  3,  etc.,  and  then  construct 
the  pressure  line  for  the  arch.  It  must,  for  sfeaiulity,  lie  within 
the  middle  third. 

To  ensure  stability  when  the  tension  in  the  rear  chain  dimin- 
ishes, or  when  the  bridge  is  unloaded,  the  arch  must  be  stable 
lyy  itself.  We  must,  therefore,  construct  the  curve  of  pressure 
f <  >r  the  arch  alone,  neglecting  the  tension  of  the  rear  chains, 
as  explained  in  the  preceding  chapter. 


CHAP.  XYI.]  THE  INVERTED  AKCH.  329 

If  this  also  passes  within  the  middle  third  of  the  arch,  as 
represented  by  the  dotted  line,  the  arch  is,  under  all  circum- 
stances, stable,  and  can  fully  resist  the  tension  of  the  rear 
chains. 

We  can  now,  finally,  so  dispose  the  joints  as  to  prevent 
sliding. 

185.  Stiffened  Suspension  System. — We  have  already  re- 
ferred to  the  methods  of  stiffening  the  cable  or  chain  so  as  to 
prevent  the  changes  of  shape  due  to  partial  loading.  Of  these 
methods,  it  only  remains  to  notice  particularly  the  last,  viz.,  by 
means  of  an  auxiliary  truss.  The  office  of  this  truss  is  to  dis- 
tribute a  partial  load  over  the  whole  length.  We  have  now  to 
investigate  the  forces  which  act  upon  the  truss. 

In  PL  27,  Fig.  107,  let  the  chain  be  acted  upon  by  the  truss 
represented  by  A  B,  which  is  called  into  action  only  by  a  par- 
tial load,  and  not  at  all  by  a  total  uniform  load.  We  can  neg- 
lect then  the  weight  of  the  truss  itself,  as  this  is  borne  by  the 
cable.  At  the  apices  3,  4,  5,  6,  7  let  us  suppose  partial  loads 
indicated  by  the  small  arrows  pointing  down.  Then,  at  every 
point  of  connection  with  the  chain,  we  have  the  reactions  1',  2', 
3',  etc.,  acting  upwards.-  Now  the  truss  must  prevent  deforma- 
tion, and  hence  these  forces  are  dependent  upon  the  form  of 
the  cable  itself.  Indeed,  if  we  take  any  point,  as  O,  as  a  pole, 
and  draw  lines  parallel  to  the  respective  sides  of  the  cable, 
these  lines  will  cut  off  upon  a  vertical  P'  these  forces.  -The 
absolute  value  of  these  forces  will,  it  is  true,  vary  according  to 
the  position  of  the  pole  assumed,  but  their  relative  proportions 
remain  always  the  same.  The  resultant  P7  of  all  these  forces 
passes  then  through  the  intersection  of  the  two  outer  sides  of 
the  catenary. 

Since  the  truss  distributes  its  load  P  upon  the  cable,  the  reac- 
tion B  at  the  right  support  is  here  zero.  The  reaction,  however, 
at  A  cannot  be  zero  unless  P  and  P'  coincide,  as  is  the  case  for 
total  uniform  load.  These,  then,  are  all  the  forces  which  are 
kept  in  equilibrium  by  the  truss.  If  P  is  given,  P'  and  the  re- 
action at  A  can  be  easily  found,  and  if  we  then  divide  P',  ac- 
cording to  the  form  of  the  chain,  into  the  portions  I/,  2',  3', 
etc.,  we  have  the  forces  at  each  apex. 

Thus  we  lay  off  to  scale  the  given  forces  3, 4,  5,  6,  7  —  P,  and 
with  a  pole  distance  any  convenient  multiple  of  the  height  of 
truss  draw  lines  to  these  points  of  division,  and  then  construct 


330  THE  INVEKTED  ARCH.  [CHAP.  XVI. 

the  corresponding  equilibrium  polygon  A  3,  4,  5,  6,  7,  B.  Pro- 
long then  the  outer  side  B  7  to  intersection  with  P',  and  draw 
the  closing  line  A  P'.  A  parallel  through  O  to  this  line  cuts 
off  from  the  force  line  P  the  reaction  at  A  and  the  cable  reac- 
tion P'. 

Now  P'  being  thus  found  and  the  form  of  cable  given,  we 
have  only  to  lay  off  P'  vertically,  draw  from  its  extremities 
lines  parallel  to  the  two  outer  sides  of  the  given  cable  arc,  and 
from  the  pole  thus  determined,  lines  parallel  to  the  other  sides 
will  give  us  the  forces  1',  2',  3',  etc.  These  when  thus  found 
we  lay  off  on  our  force  line  for  the  pole  O,  as  shown  in  the 
Fig.,  and  then  construct  the  corresponding  equilibrium  poly- 
gon A  1',  2' 9',  10',  B. 

Thus  the  vertical  ordinates  between  A  P',  P'  B  and  this 
polygon  give  us  the  moment  at  any  point  for  a  truss  acted 
upon  by  the  forces  I/,  2',  3',  etc.,  alone.  The  ordinates  between 
A  P',  P'  B  and  the  polygon  A  3,  4,  5,  6,  7,  give,  in  like  manner, 
the  moments  for  a  truss  acted  upon  by  the  forces  3,  4,  5,  6,  7, 
whose  reactions  are  A  and  P'.  The  ordinates,  then,  included 
between  both  polygons  give  us  the  moment  at  any  point  of  the 
stiffening  truss.  Thus  the  ordinate  y,  multiplied  by  the  pole 
distance,  gives  us  the  moment  in  the  truss  at  the  point  o.  If 
we  had  taken  the  pole  distance  O  equal  to  the  height  of  the 
truss,  then  these  ordinates  would  give  us  at  once  the  strain  in 
the  flanges.  We  can  thus  easily  find  the  strains  in  the  stiffen- 
ing truss  for  any  weight  or  system  of  weights  in  any  position. 

186.  Most  unfavorable  method  of  Loading. — Let  us  in- 
vestigate the  action  of  a  single  weight  P  at  any  point.  In  PI. 
27,  Fig.  109,  we  have  a  single  weight  P  acting  between  A  and 
P'. 

The  Fig.  is  nothing  more  than  a  repetition  of  Fig.  108,  only 
we  have  a  single  load  P  instead  of  a  system  of  four  loads,  and 
therefore  the  polygon  for  P  consists  only  of  two  straight  lines 
instead  of  having  as  many  angles  3,  4,  5,  etc.,  as  there  are  apex 
loads  in  the  first  case.  All  lines  have  the  same  position  as  in 
Fig.  108,  and  hence  the  construction  needs  no  further  explana- 
tion. 

We  see  at  once  from  the  Fig.  that  any  load  between  A  and 
P'  increases  the  moment  at  every  point  of  the  span  A  B,  and 
therefore  at  the  point  of  rupture  or  of  maximum  moment  also. 
So  also  for  the  shearing  force.  When,  therefore,  the  moment 


CHAP.  XVI.]  THE  INVERTED  ARCH.  331 

at  any  point,  and  the  sum  of  the  forces  between  that  point  and 
A,  is  a  maximum,  at  least  the  entire  distance  from  A  to  P' 
must  be  covered  with  the  load. 

In  Fig.  110  we  have  the  weight  P  on  the  other  side  of  the  but 
centre  P'.  The  construction  is  identical  with  Figs.  109  and  108, 
the  position  and  the  direction  of  action  of  the  forces  is  now 
different.  Since  the  resultants  A  and  P'  now  lie  on  the  same 
side  of  P,  A  and  P'  act  in  opposite  directions,  and  since  P' 
must  still  act  upwards,  A  must  act  downwards.  In  the  neigh- 
borhood of  5'  the  moment  is  zero.  Between  this  point  and  B 
the  moments  have  the  same  signification  as  in  Fig.  109 ;  on 
the  other  side  the  moments  have  then  a  different  siom.  In 

O 

order,  then,  that  the  moment  at  5 '  shall  be  a  maximum,  the  load 
must  cover  the  length  from  A  to  P,  this  last  point  being  the 
point  at  which  a  load  causes  no  moment  in  5';  for  if  any 
point  between  A  and  P  were  not  loaded,  as  we  have  seen,  a 
load  at  that  point  would  increase  the  moment  at  5'.  A  load 
beyond  P,  however,  would  diminish  the  moment  at  5'. 

The  above  holds  good  for  every  point  between  A  and  P',  and 
therefore  for  the  point  of  rupture  or  of  maximum  moment  it- 
self. In  order  that  this  maximum  moment  can  be  no  more 
increased,  the  load  must  extend  from  A  beyond  the  centre  to 
that  point  at  which  a  load  being  placed  causes  no  moment  at 
the  cross-section  of  rupture. 

As  for  the  shearing  force,  at  the  end  A  it  will  evidently  be 
greatest  for  load  from  A  to  P',  or  over  the  half  span,  since 
every  load  on  the  other  side  of  P'  diminishes  this  reaction. 
Hence  we  have  the  following  principles  established  : 

The  moment  cut  any  Gross-section  of  the  stiffening  truss  is  a 
maximum,  when  the  load  reaches  from  the  nearest  end  beyond 
the  centre  to  a  point  for  which  the  moment  at  this  cross-section 
is  zero. 

The  above  condition  holds  good,  therefore,  for  the  maximum 
of  all  the  maximum  moments,  or  for  the  cross-section  of  rup- 
ture itself. 

The  maximum  shearing  force  is  at  one  end  of  the  truss 
when  the  adjacent  half  span  is  loaded. 

If  the  arc  is  unsymmetrical,  we  must  understand  by  "  half 
span  "  the  distance  between  the  end  and  vertical  through  the 
intersection  of  the  outer  arc  ends  produced. 

1ST.    Example. — As  an  illustration  of  the  above  principles, 


332  THE    INVERTED    ARCH.  [CHAP.  XVI. 

let  us  take  the  structure  represented  in  PL  28,  Fig.  111.  Span, 
60  ft. ;  depth  of  truss  and  panel  length,  5  ft.  Scale,  10  ft.  to  an 
inch.  We  suppose  the  live  load  to  be  2  tons  per  ft.,  giving 
thus  10  tons  for  each  lower  apex,  and  take  the  scale  of  force  50 
tons  per  inch. 

On  the  left  we  have  laid  off  the  force  lines  for  the  loads  2,  3, 
4,  5,  6  and  7  to  11,  and  have  taken  the  poles  'for  each,  so  that 
the  first  lines  are  all  parallel  to  each  other  and  to  the  first  link 
of  the  cable ;  the  common  pole  distance  being  2  j-  times  the 
height  of  trass,  or  1.5  inches.  The  moment  scale  is  then 
1.5  x  10  x  50  =  750  ft.  tons  per  inch.  Since  the  full  load  is 
entirely  supported  by  the  cable,  we  have  only  to  investigate  the 
effect  of  the  live  load  upon  the  truss. 

Precisely  as  in  Fig.  108,  we  construct  the  polygons  for  forces 
2-11,  3-11,  4-11,  etc.,  and  draw  the  closing  lines  as  indicated 
by  the  broken  lines  radiating  from  the  centre  O.  Parallels  to 
these  from  the  poles  cut  off  from  the  force  lines  the  end  and 
chain  reactions.  The  upper  portions  are  the  chain  reactions, 
the  lower  the  reactions  at  the  right  end  for  the  loads  2-11, 
3-11,  etc. 

Now  we  have  to  divide  these  chain  reactions  into  as  many 
parts  as  there  are  load  apices  by  lines  parallel  to  the  sides  of 
the  chain.  This  we  have  done  by  drawing  two  lines  parallel  to 
the  two  chain  ends,  inserting  the  chain  reactions  between  these 
lines,  and  then  drawing  parallels  to  the  chain  sides.  If,  as  in 
this  case,  the  curve  of  the  chain  is  a  parabola,  these  reactions 
are  divided  into  11  equal  parts.  If  the  chain  has  any  other 
form,  the  parallels  to  the  chain  sides  determine  the  relative 
lengths  of  these  portions. 

It  will  only  be  found  necessary  to  construct  the  moment 
polygons  for  4,  5  and  6-11 ;  the  other  polygons  already  drawn 
are  necessary  for  the  determination  of  the  shearing  forces  only. 

Thus,  on  the  force  line  for  loads  4  to  11  we  can  now  lay  off 
the  11  equal  parts  just  found,  into  which  the  chain  reaction  is 
divided.  So  for  5-11  and  6-11.  These  portions  we  have  indi- 
cated by  Roman  numerals.  We  can  now  draw  the  correspond- 
ing polygons  precisely  as  in  Fig.  108,  which  are  indicated  also 
by  Roman  numerals. 

It  is  then  easy  with  the  dividers  to  pick  out  the  maximum 
moment  at  any  apex.  These  moments,  laid  off  as  below,  give 
the  curve  of  moments  for  the  truss,  which  being  scaled  off  and 


CHAP.  XVI.]  THE  INVERTED  ARCH.  333 

divided  by  the  depth  of  truss,  give  at  once  the  strains  in  the 
flanges.  Since  the  moment  scale  is  750  tons  per  inch  and  the 
depth  of  truss  5  ft.,  the  moment  ordinates  scaled  off  at  150  tons 
per  inch  will  give  at  once  the  strains  in  the  flanges,  without 
division. 

For  the  shearing  forces,  we  know  from  the  preceding  that 
the  maximum,  reaction  at  right  end  is  for  loads  6-11.  This 
reaction  we  have  already  found  in  the  corresponding  force  line 
by  means  of  the  closing  line  already  drawn.  We  lay  it  off  then 
right  and  left,  half-way  between  the  ends  and  first  apex,  that 
being  the  effective  length  of  load,  the  two  half-end  panels  rest- 
ing directly  upon  the  abutments. 

The  maximum  shear  at  any  point  is  evidently  when  the  load 
reaches  from  right  support  to  that  point,  and  is  equal  to  the 
sum  of  the  chain  reactions  at  the  unloaded  apices.  Thus,  max- 
imum shear  at  3  is  equal  to  the  interval  I II  for  the  line 
3-11;  at  4,  I  III  for  line  4-11;  at  5,  I IV  for  line  5-11; 
and  at  6,  I V  for  line  6-11.  Laying  off  the  shear  at  6,  we 
can  draw  the  line  6-11,  as  indicated  in  the  diagram,  and  thus 
determine  the  shear  at  2.  This  we  cannot  find,  as  above,  for  3, 
4,  etc.,  as  for  the  load  2-11 ;  owing  to  the  shape  of  the  chain 
as  represented,  there  is  no  upward  reaction  at  1,  as  there  is  no 
angle  of  the  chain  at  6. 

The  shear  diagram  is,  of  course,  symmetrical  on  each  side  of 
the  centre.  We  can  therefore  construct  it  as  represented,  and 
then  the  determination  of  the  strain  in  the  diagonals  is  easy. 
We  have  only  to  multiply  the  shear  at  any  apex  by  the  secant 
of  the  angle  which  the  diagonals  make  with  the  vertical.  This 
we  may  do  by  properly  changing  the  scale  at  once,  and  thus 
scale  off  the  strains  directly. 

188.  Analytical  Determination  of  the  Forces  acting 
upon  the  stiffening  Tru§s. — Assuming  that  the  truss  distrib- 
utes the  partial  loading  uniformly  over  the  whole  arc,  we  may 
deduce  very  simple  formulae  for  the  forces  acting  upon  the 
truss.  As  we  have  already  seen,  for  a  maximum  moment  at 
any  point,  the  load  must  always  extend  out  from  one  end. 

Let  us  represent,  then,  the  ratio  of  the  loaded  part  from  left 
to  the  whole  span  by  k. 

Let  the  entire  span  be  2  Z,  then  the  loaded  portion  is  2  k  I. 
Let  m  be  the  load  per  unit  of  length ;  then  the  whole  load 
[Fig.  108]. 


334  THE  INVERTED  ARCH.  [CHAP.  XVI. 

The  distance  of  P  from  the  left  is  then  half  the  loaded  por- 
tion, or  k  I.  Its  distance  from  P',  which  acts  at  the  centre  of 
the  span,  is  I  (1  —  &). 

Hence  we  have  for  the  left  reaction  A 
Axl  =  Pl(l-k)   or  A  =  -p(l-fy 
Also  P'l  =  P  x  Tel  or  P'  =  2k*l 

The  chain  reaction  per  unit  of  length  is  then 


Now  let  x  be  the  distance  to  the  point  of  maximum  moment. 
Now  since  at  this  point  the  shear  must  be  zero,  the  weight 
of  the  portion  x  must  be  equal  to  A  (Art.  38). 
We  have  then 

Ax  -  ~  =  P  (kl  -  x)  =  2Jclm(Jcl  -  x), 

Zi 

whence,  by  substituting  the  value  of  A, 

x  = ^. 

1  +  k 

But  the  maximum  moment  is  Ax — -  =  -— ,  and  there- 
fore, substituting  the  value  of  x  above, 

_    _  fC       \  X      "^     fC )  /-\     70 

M  max.  =  — j =—£ .  2 1  m. 

1  +  K 

This  becomes  a  maximum  for  1  —  k  —  %?  =  0,   or  for 
Jc  =  i  1/5  -  i  =  0.618034. 

Therefore,  the  greatest  moment  occurs  when  0.62  of  the  span 
is  covered  with  the  load. 

We  have  then  the 

Length  of  the  loaded  portion,  =  2  k  I  =  0.61803  x  2  .1 
Reaction,  A  =  2  & Z  ra  (1  -  &)  =  (  4/5  —  2) 2Zm  =  0.23607. 
Chain  reaction,  P'  =  *2&?lm  =  (J-  —  ^  1/5)2 lm  =  0.38196. 
Load  per   unit  in  loaded  portion,  or  the  difference    between 
the  load  m  and  the  chain  reaction  m  1£  per  unit  of  length 
=  m  (1  -  &')  =  i  ( 1/5  -  1)  m  ?=  0.61803  m. 

The  distance  of  the  point  of  maximum  moment  is 

A 


x  =  -          =  (t  -  4  ^5)  2  ^  =  0.38196  .  2  Z  = 


CHAP.  XVI.]  THE  INVESTED  AECH.  335 

The  maximum  moment  itself  is 


-  TO\  -4  7  u-  .  *  v  \  v       '         "  ---   *.     F     v          *  *  v  

2  m  (1  —  F)          1  +  & 

For  a  simple  girder  uniformly  loaded,  the  maximum  moment! 
is  \p  1?.  The  maximum  moment  is  then  reduced  from  \  to 
0.18,  or  to  about  £d,  or  is  -f^ths  the  maximum  moment  for  a 
simple  girder  of  same  span  and  load* 

If  we  represent  the  dead  load  by  p,  then,  since  the  stiffening 
truss  sustains  only  the  moving  load,  we  have 

0.18  ml?      _     36      m  \ m  (0.6  If 

That  is,  the  maximum  moment  in  the  stiffening  truss  is  the 
same  as  for  a  simple  girder  of -f^ths  the  span,  loaded  only  with 
the  moving  load. 

189.  Summary, — The  reaction  at  the  end  abutment  and  the 
chain  reaction  at  each  apex  having  been  found,  as  above,  for 
any  given  load,  we  might  have  found  the  strains  in  every 
piece  by  the  method  of  Arts.  8-13.  This  would,  however,  in 
this  case  have  proved  long  and  tedious.  The  construction  of 
the  curve  of  maximum  moments  and  shear  is  preferable. 

We  can  therefore  readily  determine  the  strains  in  such  a 
combination  as  that  represented  in  Fig.  111.  We  have  already, 
Arts.  90-94,  given  practical  and  simple  methods  for  the  deter- 
mination of  the  strains  in  braced  arches  of  the  usual  forms  of 
construction. 

It  will  be  observed  that  it  is  by  no  means  necessary  that  the 
arrangement  of  bracing  and  flanges  should  be  the  same  as  that 
shown  in  Figs.  90  and  94. 

Thus  we  may  treat  the  arch  represented  in  Fig.  5  (c)  accord- 
ing  to  Art.  158.  as  hinged  at  both  abutments  and  crown,  or, 
making  the  lower  flange  continuous  at  the  crown,  we  may  find 
the  resultant  pressures  at  the  abutments  by  Art.  159,  and  then 
follow  these  pressures  through  precisely  as  shown  in  the  Fig. 

The  combination  of  Fig.  Ill  being  of  considerable  impor- 
tance, as  the  more  usual  form  of  construction  of  suspension 
bridges,  and  not  falling  under  our  classification  of  "  braced 
arches,"  we  have  considered  it  desirable  to  discuss  it  somewhat 

*  Rankine  gives  -/7-ths  for  a  girder  whose  ends  are  fixed,  the  greatest  mo- 
ment occurring  for  a  load  over  Jds  the  span. 


336  THE  INVERTED  ARCH  [CHAP.  XIV. 

fully.  A  better  form  of  construction  is  that  shown  in  Fig. 
106,  which  is  perfectly  rigid,  and  the  strains  in  which  are 
easily  found  by  Art.  158  or  159,  according  as  we  hinge  it  in 
the  centre  or  not. 


Reviewing  now  the  preceding,  we  see  that  the  graphical 
method,  as  here  developed,  furnishes  us  with  a  simple,  accurate 
and  practical  solution  of  nearly  every  class  of  structure  occur- 
ring in  the  practice  of  the  engineer  or  builder.  In  our  first 
chapter  we  have  a  method  by  the  resolution  of  forces  applica- 
ble to  any  framed  structure,  however  irregular  or  unsymmetri- 
cal,  provided  only  there  are  no  moments  at  the  ends  to  be 
determined. 

In  Art.  125  we  have  explained  fully  the  application  of  the 
method  for  this  case  also,  when  these  moments  are  known,  and 
in  Chaps.  VIII.  to  XIV.  inclusive  we  have  given  practical  con- 
structions for  the  determination  of  these  moments  for  all  the 
important  classes  of  structures  in  which  this  condition  occurs, 
such  as  the  continuous  girder,  braced  arches,  etc. 

When  the  structure  is  not  framed,  or  composed  of  pieces  the 
strains  in  which  can  be  definitely  determined,  we  have  the 
method  of  moments  of  Chap.  V.,  which,  as  we  have  seen,  may 
be  extended  so  as  to  completely  solve  the  difficult  case  of  the 
continuous  girder,  and  which  may,  of  course,  be  applied  to 
framed  structures  also,  as  illustrated  in  Fig.  Ill  (Art.  187)  in 
the  case  just  discussed.  Thus  we  have  two  distinct  graphical 
methods  by  which  our  results  may  be  checked.  The  first- 
method  includes  a  great  variety  of  the  most  important  and 
usual  structures,  such  as  bridge  girders,  roof  trusses,  cranes, 
etc.,  and  in  view  of  its  ease  and  accuracy  will  undoubtedly  be 
found  of  great  service  by  the  engineer  and  architect.  The 
second  method  has  important  mechanical  applications,  as  no- 
ticed in  Art.  41 ;  and  aside  from  these,  and  its  application  to 
structures  having  end  moments,  such  as  the  continuous  girder, 
etc.,  furnishes  us  with  ready  determinations  of  the  centre  of 
gravity  of  areas  (Chap.  III.),  the  moment  of  inertia  of  areas 
(Chap.  VI.),  and  also  gives  us  a  very  complete  solution  of  the 
stone  arch  (Chap.  XV.). 

We  have  also  the  analogous  methods  of  calculation,  viz., 


CHAP.  XVI.]  THE  INVEKTED  ARCH.  337 

both  by  resolution  of  forces  and  by  moments  (Arts.  9  and  16 
of  Appendix).  The  latter  being  so  general  and  simple  in  its 
application,  we  have  not  felt  justified  in  leaving  it  entirely  out 
of  sight,  and  in  those  cases  where  it  seemed  of  especial  service, 
or  assisted  the  graphical  solution,  we  have  illustrated  it  more  or 
less  fully  (Chap.  XII.).  In  this  latter  chap,  we  have  also  given 
constructions  as  well  as  formulae,  and  developed  principles 
which,  it  is  believed,  render  possible,  for  the  first  time,  the  com- 
plete and  accurate  solution  of  the  important  .case  of  the  "  draw 
xpan."  (Arts.  118-121.) 

The  formulae  of  Chap.  XIII.  in  connection  with  the  method 
of  calculation  by  moments,  render  the  calculation  of  the  con- 
tinuous girder  generally  as  simple,  and  but  little  more  tedious 
than  for  the  simple  girder  itself.  Whatever  may  be  thought 
of  the  advantages  or  disadvantages  of  this  class  of  structures 
by  engineers  generally,  it  is  at  least  time  that  such  structures 
as  draws  or  pivot  spans  should  be  calculated  under  suppositions 
which  approach  somewhat  more  nearly  the  actual  case  than  is 
at  present  the  practice.  As  to  the  relative  economy  of  con- 
tinuous girders,  we  have  endeavored  to  enforce  the  fact  that 
the  saving  over  the  simple  girder  is  from  15  to  20  and  even  50 
per  cent.  We  give  in  the  Appendix  a  tabular  comparison  of  a 
few  cases  sufficient  to  show  the  point  beyond  dispute,  and  any 
one  may  easily  add  to  the  list,  or  verify  the  calculations. 

The  "  graphical  arithmetic,"  as  it  might  be  called,  such  as 
graphical  addition,  subtraction,  multiplication,  division,  extrac- 
tion of  roots,  determination  and  transformation  of  areas,  etc., 
we  have  entirely  omitted  in  the  present  work,  judging  it  of  but 
little  practical  value,  except  in  rare  cases,  when  we  have  ex- 
plained the  necessary  constructions  as  they  occur,  and  unneces- 
sary for  the  development  of  the  graphical  method  proper.  [See 
Chap.  IV.  of  Introduction.] 
22 


APPENDIX. 


GRAPHICAL    STATICS. 


A.  JAY  DUBOIS. 


APPENDIX. 


NOTE   TO    CHAPTER   VIII.    OF    THE    INTRODUCTION  -  UPON    THE 
MODERN    GEOMETRY.* 

IT  is  to  be  regretted  that,  notwithstanding  its  beauty  of  form, 
simplicity,  and  many  happy  applications  in  the  technical  and 
natural  sciences,  the  Modern  Geometry  is  yet  hardly  known, 
scarcely  by  name  even,  in  our  schools  and  colleges. 

The  work  of  Gillespie  upon  Land  Surveying,  already  cited  in 
the  Introduction,  and  a  treatise  on  Elementary  Geometry  by 
William  Chauvenet  (Phil.,  1871),  are  the  only  ones  which 
occur  to  us  in  this  connection. 

It  has  already  been  stated  that  the  modern  orj)ure  geometry 
of  space  differs  essentially  from  the  ancient,  and  from  analytical 
geometry,  in  that  it  makes  no  use  of  the  idea  of  measure  ny  r>f 
We  find  in  it  no  mention  of  the  bisection 


of  lines,  of  right  angles  and  perpendiculars,  of  areas,  etc.,  any 
more  than  of  trigonometrical  quantities,  or  of  the  analytical 
equations  of  lines.  We  have  nothing  to  do  with  right-angled, 
equilateral,  or  equiangular  triangles,  with  the  rectangle,  regular 
polygon,  or  circle,  pxcept  in  a  supplementary  manner.  So  also 
for  the  centre,  axes,  and  foci  of  the  so-called  curves  of  the  second 
order,  or  the  conic  sections. 

On  the  contrary,  we  obtain  much  more  general  and  compre- 
hensive properties  of  these  curves  than  those  to  which  most 
text-books  upon  analytical  geometry  are  limited. 

A  new  path  is  thus  opened  to  the  conic  sections,  without  the 
aid  of  the  circular  cone,  after  the  manner  of  the  ancients,  or  of 
the  equations  of  analytical  geometry. 

As  a  direct  consequence,  the  principles  and  problems  of  the 
modern  geometry  are  of  great  generality  and  comprehensive- 

*  The  following-  remarks  and  illustrations  are  taken  from  the  Geometrie  der 
Lage,  by  Reye.  Hannover,  1866. 


342        NOTE  TO  CHAP.  VIH.  OF  THE  INTKODUCTION.  [APPENDIX. 

ness.  Thus  the  most  important  of  those  properties  of  the  conic 
sections  which  are  proved  in  text-books  of  analytical  geometry 
are  but  special  cases  of  its  principles.  A  few  particular  ex- 
amples taken  from  the  Geometrie  der  Lage,  by  Reye,  which 
could  not  well  have  been  inserted  in  the  Introduction  to  this 
work,  will  best  explain  and  illustrate  our  general  remarks — the 
more  so  as  these  examples  are  of  special  interest  and  value  to 
the  engineer. 

It  is  a  problem  of  frequent  occurrence  in  surveying  to  pass  a 
line  through  the  inaccessible  and  invisible  point  of  intersection 
of  two  given  lines.  The  Geometry  of  Measure,  or  ancient 
geometry,  gives  us  any  required  number  of  points  upon  this  line 
by  the  aid  of  the  principle,  that  the  distances  cut  off  from  par- 
allel lines  by  any  three  lines  meeting  in  a  common  point  are 
proportional.  The__Geometry  of  Position  furnishes  us  with  a 
simpler  solution. 

FIG.  1. 


Thus  the  two  lines  «,  5  being  given  [Fig.  1.],  we  have  sim- 
ply to  choose  any  point  we  please,  as  P.  From  this  point  draw 
any  number  ;of  lines  desired,  in  any  direction  intersecting  the 
given  lines.  Now,  in  any  quadrilateral  which  any  two  of  these 
lines  form  with  the  two  given  lines  a  and  &,  we  have  simply  to 
draw  the  diagonals.  The  intersections  of  all  these  diagonals 
lie  in  the  same  straight  line  passing  through  the  intersection  A 
of  the  two  given  lines,  and  therefore  determine  the  line  re- 
quired. Observe  that  the  construction  is  entirely  independent 
of  all  metrical  relations,  and  depends  solely  upon  the  relative 
position  of  the  two  given  lines. 

Again  :  If  we  take  upon  any  straight  line  three  points,  A,  B 
and  C  [Fig.  2.],  and  construct  any  quadrilateral,  two  opposite 
sides  of  which  pass  through  A,  one  diagonal  through  B,  and  the 


APPENDIX.] 


THE  MODERN  GEOMETRY. 


343 


other  two  opposite  sides  through  C,  then  will  the  other  diagonal 
intersect  the  line  in  a  point  D,  which  for  the  same  three  points, 


FIG.  2. 


A  B  C  D 

A,  B  and  C,  is  always  the  same  for  every  possible  construction. 
Moreover,  these  four  points,  A,  B,  C  and  D,  are  always  harmonic 
points,  so  that  D  is  harmonically  separated  from  B  by  the 
points  A  and  C.  Thus,  A  B  :  B  C  ; ;  A  D  :  C  D.  This  construc- 
tion may  also  be  applied  in  surveying,  as  in  passing  around  an 
obstruction,  as  a  wood,  etc.,  into  the  same  line  again. 

Again :  We  may  notice  the  following  principle  concerning 
the  triangle  [Fig.  3]  ; 

FIG.  3. 


u. 


If  two  triangles,  ABC  and  A!  Bt  C1?  are  so  situated  that  the 
lines  joining  corresponding  angles,  as  A  A1?  B  B1?  C  C1;  meet  in 
a  common  point  S,  then  will  the  intersections  of  corresponding 
sides,  as  A  C  and  Ax  Gl9  A  B  and  At  B1?  B  C  and  B!  C1?  meet  in  a 
common  line,  as  u  u.  The  inverse  also,  of  course,  holds  good : 
that  if  the  sides  intersect  on  a  line,  the  lines  through  the  angles 
intersect  in  a  point. 

Another  series  of  principles  are  connected  with  the  curves  of 
the  second  order,  or  conic  sections.  From  analytical  geometry, 


344  NOTE   TO    CHAP.  VIII.  OF   THE    INTRODUCTION.    [APPENDIX. 

as  is  well  known,  a  curve  of  the  second  order  is  completely  de- 
termined by  five  points  or  five  tangents.  But  the  length  of  the 
calculation  or  construction  of  a  curve  thus  determined  is  also 
well  known.  The  geometry  of  position,  however,  proves  two 
very  important  principles,  which  render  it  easy  to  construct  to 
the  five  given  points  or  tangents  any  number  of  new  points  or 
tangents,  and  thus  quickly  draw  the  curve  itself.  The  reader 
already  acquainted  with  these  principles  will  also  probably  re- 
member how  much  auxiliary  demonstration  their  proof  in  the 
analytical  geometry  requires.  The  first  of  these,  due  to  Pascal, 
is,  that  the  three  pairs  of  opposite  sides  of  a  hexagon  inscribed 
within  a  conic  section  intersect  upon  a  straight  line.  The 
second,  due  to  Brianchon^  is,  that  the  three  principal  diagonals 
of  the  circumscribing  hexagon,  which  unite  every  pair  of  oppo- 
site angles,  intersect  in  one  and  the  same  point.  Both  prin- 
ciples are  easily  deduced  from  the  circle.  It  will  be  observed 
that  they  are  independent  of  the  relative  dimensions,  centre, 
axes,  and  foci  of  the  curves.  For  this  very  reason  they  are  of 
the  greatest  generality  and  significance,  so  that  an  entire  theory 
of  the  conic  sections  can  be  based  upon  them.  Thus  Pascal's 
principle  solves  the  important  problem  of  tangent  construction 
from  a  given  point,  even  when  the  curve  is  given  by  five  points 
only,  without  completely  constructing  it. 

This  problem  of  tangent  construction  to  curves  of  the  second 
order  can  in  many  cases  be  solved  by  the  aid  of  a  principle 
which  expresses  one  of  the  most  important  properties  of  the 
conic  sections,  but  which,  nevertheless,  is  seldom  found  in  text- 
books upon  analytical  geometry,  because  its  analytical  proof  is 
somewhat  complicated,  and  little  suited  to  set  forth  the  property 
in  its  proper  light. 

For  example  :  If  through  a  point  A  [Fig.  4]  in  the  plane  of 
but  not  lying  upon  a  curve  of  the  second  order,  we  draw  se- 
cants, every  two  secants  determine  four  points,  as  K,  L,  M,  N, 
upon  the  curve.  Any  two  lines  joining  these  four  points,  as 
L  M  and  K  N  or  K  M  and  Li  l\r,  intersect  in  a  point  of  a 
straight  line  a  a,  which  is  the  polar  of  the  given  point  A  ;  that 
is,  which  intersects  the  curve  in  the  two  points  of  tangency 
G  G.  Thus  the  lines  through  A  and  the  intersections  of  a  a 
with  the  curve  are  the  tangents  to  the  curve  through  A.  If 
the  point  A  were  within  the  curve,  this  line  a  a  would  not  in- 
tersect it.  This  construction  can  be  used  in  order  to  draw 


APPENDIX.]          THE  MODEEN  GEOMETRY.  345 

through  a  given  point  tangents  to  a  conic  section  by  the  sim- 
ple application  of  straight  lines.     Upon  every  secant  through 

FIG.  4. 
\L 


A,  moreover,  there  are  four  remarkable  points,  viz. :  the  point 
A  itself,  the  first  intersection  B  with  the  curve,  the  intersection 
with  the  polar,  and,  finally,  the  second  intersection  D  with 
the  curve.  These  four  points  are  harmonic  points,  and  the 
polar  a  a  contains,  then,  every  point  which  is  harmonically 
separated  from  A  by  the  two  curve  points.  The  important, 
principles  relating  to  the  centre  and  conjugate  diameter  of 
conic  sections  are  merely  special  cases  of  the  above  important 
principles.  These  last  can  be  easily  extended  to  surfaces  of 
the  second  order,  as  the  intersection  of  these  by  a  plane  is,  in 
general,  a  curve  of  the  second  order. 

From  these  few  examples,  which  might  be  indefinitely  multi- 
plied, it  may  easily  be  seen  how  very  different,  but  not  less  im- 
portant than  those  of  analytical  geometry,  are  the  theorems  of  the 
geometry  of  position.  Thus  the  latter  are  generally  proved  by 
aid  of  the  angle  which  the  tangents  make  with  the  line  through 
the  focus,  or  by  the  distances  cut  off  from  the  axes — that  is,  by 
metrical  relations.  We  refer,  of  course,  to  the  elements  of 
analytical  geometry  as  contained  in  most  text-books,  and  not  to 
those  most  fruitful  and  later  methods  .whose  existence  are 
chiefly  due  to  the  sagacity  of  Plucker  (Introduction,  VIII.). 


346  NOTE    TO    CHAP.    I.  [APPENDIX. 


NOTE  TO  CHAPTEE  I. 

1.  The  method  by  the  resolution  of  forces  developed   in 
Chapter  I.  is  so  simple  and  easy  of  application,  and  its  principles 
are  so  few  and  self-evident,  that  we  have  not  considered  it  ad- 
visable to  tax  the  patience  of  the  reader  by  any  great  variety 
of  practical  applications.     A  large  number  of  such  applica- 
tions are  to   be  found  in  a  most  excellent  little  treatise  by 
Robert  H.  Bow,  entitled  The  Economics  of  Construction  in 
Relation  to  Framed  Structures.     There  are,  however,  a  few 
important  practical  points- of  detail,  and  a  few  general  consid- 
erations, which  we  think  it  well  to  notice  here,  and  to  which, 
in  illustration   of  the  remarks  in  Chap.  I.,  the  reader  will  do 
well  to  attend. 

2.  In   PL  1,  Fig.  I.  (Appendix),  we  have  represented  the 
"Bent  Crane  "  given  by  Stoney  in  his   Theory  of  Strains,  p. 
121,  Art.  200. 

We  assume  the  following  method  of  notation.  Let  all  that 
space  above  the  Fig.  be  indicated  by  X,  and  all  that  space 
below  by  Y,  and  the  triangular  spaces  enclosed  by  the  flanges 
and  diagonals  by  the  numbers  1,  2,  3,  4,  etc.  The  first  upper 
flange  is  then  denoted  by  X  2,  the  second  by  X  4,  and  so  on. 
So  also  the  first  lower  flange  is  Y  1,  the  next  Y  3,  etc.  The 
first  diagonal  is  then  X  1,  the  next  1  2,  the  next  2  3,  etc.* 

The  flanges  are  equidistant,  forming  quadrants  of  two  cir- 
cles whose  radii  are  respectively  20  and  24  feet.  The  inner 
flange  is  divided  into  four  equal  bays,  on  which  stand  isosceles 
triangles,  and  a  weight  of  10  tons  is  suspended  from  the  peak. 
The  scale  for  this  and  all  the  Figs,  of  PI.  I.  is  20  tons  to  an  incji 
and  10  feet  to  an  inch.  Laying  off,  then,  the  weight  X  Y  =  10 
tons,  we  form,  according  to  the  method  of  Chapter  I.,  the  strain 
diagram.  It  will  be  seen  at  once  that  all  the  lower  flanges,  Y  1, 
Y  3,  etc.,  radiate  from  Y,  all  the  upper  flanges,  X  2,  X  4,  etc., 
from  X,  and  everywhere  the  letters  in  the  one  diagram  indi- 
cate the  corresponding  pieces  in  the  other. 

*  For  this  very  elegant  niethod  of  notation,  we  are  indebted  to  the  work  of 
B.  H.  Bow,  above  alluded  to. 


APPENDIX.]  NOTE   TO   CHAP.  I.  347 

"We  can  now  at  once  take  off  the  strains  to  scale  in  the  vari- 
ous pieces. 

A  comparison  of  our  method  with  that  given  by  Stoney  for 
the  same  case  will  be  instructive,  as  illustrating  the  compara- 
tive merits  of  the  two. 

3.  Character  of  the   Strain*   in  the  Pieces. — One  of  the 
most  important  points  of  our  method  is  the  ease  and  certainty 
with  which  the  character  of  the  strains  in  the  pieces  may  be 
determined.    We  have  only,  as  detailed  at  length  in  Chapter  1., 
to  follow  round  any  closed  polygon  in  the  direction  of  the 
forces,  and  then  refer  back  to  that  apex  of  the  frame  where  for 
the  moment  we  may  happen  to  be. 

Thus  for  the  peak,  since  we  know  that  the  weight  acts  down, 
we  follow  down  from  X  to  Y,  and  then  from  Y  to  1,  and  1 
back  to  X.  Referring  back  now  to  the  frame,  and  remember- 
ing that  a  force  acting  away  from  the  apex  means  tension,  and 
towards,  compression,  we  have  at  once  Y  1  compression  and 
X  1  tension. 

Now  for  apex  a,  since  X  1  is  tension,  with  respect  to  this  new 
apex,  it  must  act  away.  We  go  round  then  from  X  to  1,  1  to 
2,  and  2  back  to  X,  and  then,  referring  these  directions  to  the 
corresponding  pieces  meeting  at  &,  we  have  1  2  compression 
and  X  2  tension. 

We  find  thus  all  the  outer  flanges  in  tension,  as  evidently 
should  by  simple  inspection  be  the  case.  Also  all  the  inner 
flanges  compression.  As  for  the  diagonals,  they  alternate,  the 
first  being  tension,  the  next  compression,  until  we  arrive* at  4  5, 
which  we  find  to  be  also  compression. 

A  glance  at  the  strain  diagram  shows  how  this  comes  about. 
The  line  X  4  crosses  Y  5,  and  thus  gives  us  a  reverse  direction 
for  4  5. 

In  such  a  simple  structure  as  the  present,  the  character  of  the 
strains  would  present  no  especial  difficulty  in  any  case ;  but  in 
more  complicated  ones,  the  aid  of  such  a  simple  and  sure  crite- 
rion as  the  above  is  indispensable,  and  we  have  been  thus  even 
prolix  upon  this  point,  the  more  so  as  it  is  not  so  much  as 
alluded  to,  as  far  as  we  are  aware,  in  those  few  works  which 
notice  the  above  method  at  all. 

4.  There  are  other  points  which  we  may  here  illustrate  by 
our  Fig. 

According  to  our  first  principle  (Art.  3,  Chapter  I.),  when 


348  NOTE   TO    CHAP.  I.  [APPENDIX. 

any  number  of  forces  are  in  equilibrium,  the  force  polygon  is 
closed.  Inversely,  then,  a  closed  force  polygon  indicates  forces 
which,  if  applied  at  a  common  point,  would  hold  each  other  in 
equilibrium. 

Thus  Y  3,  34,  X  4,  and  the  weight  are,  or  would  be,  if  all 
applied  at  a  common  point,  in  equilibrium.  This  we  see 
directly  from  the  Fig.  Thus  we  know  that  when  any  num- 
ber of  forces  are  in  equilibrium,  the  algebraic  sums  of  their 
vertical  and  horizontal  components  must  be  zero,  otherwise 
there  must,  of  course,  be  motion.  Now  the  vertical  component 
of  Y  3  plus  that  of  3  4  minus  that  of  X  4  is  exactly  equal  and 
opposed  to  the  weight,  while  the  horizontal  component  of  Y  3 
plus  that  of  3  4  is  equal  and  opposed  to  that  of  X  4,  and  there 
is  then  equilibrium. 

Again,  according  to  the  principle  of  Art.  5,  Chap.  I.,  any 
line,  as  the  one  joining  2  and  6  (broken  line  in  Fig.),  is  the 
resultant  of  X  2  and  X  6,  as  also  of  2  3,  3  4,  4  5  and  5  6. 

The  Fig.  also  well  illustrates  the  points  to  be  avoided  in 
making  a  strain  diagram,  already  alluded  to  in  Art.  13,  Chap. 
I.  The  scale  to  which  the  frame  is  taken  is  here  altogether 
out  of  proportion  to  the  scale  of  force.  The  first  should  be 
increased  or  the  second  diminished,  or  both.  The  present 
length  of  the  diagonals  and  flanges  is  inadequate  to  give 
with  sufficient  accuracy  the  directions  of  strain  lines  of  such 
length. 

Nevertheless  we  have  experienced  no  difficulty  in  checking 
to  tenths  of  a  ton  the  results  given  by  Stoney  for  this  structure. 

5.  In  PI.  1,  Fig.  II.,  we  have  represented  a  roof  truss,  span 
30  ft.,  rise  8  ft.,  camber  1  ft. ;  and  the  strain  diagram  illustrates 
in  its  two  symmetrical  halves  (one  full,  the  other  dotted)  the 
remarks  of  Art.  13,  Chap.  I.,  upon  the  check  which  in  such  cases 
our  method  furnishes  of  its  accuracy. 

We  lay  off  the  weights  1,  2,  3,  4,  5,  and  then  the  reactions  at 
A  and  B,  which  should  bring  us  back  to  the  point  of  beginning, 
and  thus  complete  the  force  polygon.  The  strains  are  then  easily 
found,  and  the  two  halves  should  be  perfectly  symmetrical,  and 
give  the  same  results. 

In  Fig.  III.  we  have  given  another  form  of  truss  with  strain 
diagram,  the  other  half  of  which  the  reader  can  complete  and 
letter  for  himself. 

6.  In  Fig.  IY.  we  have  a  form  called  the  French  roof  truss 


APPENDIX.]  NOTE   TO   OH  A  P.  I.  349 

and  two  strain  diagrams — the  larger  for  vertical  reactions,  the 
smaller  for  inclined  reactions. 

This  last  brings  out  the  force  polygon  in  perhaps  a  clearer 
shape  than  before.  The  weights  1  to  Y  being  laid  off  down- 
wards, the  two  reactions  must  always  bring  us  back  to  the 
starting-point,  and  thus  close  the  polygon — in  this  case  a  trian- 
gle, in  the  preceding  case  a  straight  line,  and  in  the  case  of  Fig. 
6,  Art.  10.,  Chap.  I.,  a  true  polygon.  Both  strain  diagrams 
illustrate  the  check  we  have  upon  the  accuracy  of  the  work. 
The  second  half  should  be  perfectly  symmetrical  with  the  first, 
and  the  lines  Y  k  and  points  Jc  in  each  should  coincide. 

We  have  here  also  to  notice  a  point  which  in  roof  trusses  is 
of  frequent  occurrence,  and  may,  if  not  noticed,  cause  diffi- 
culty. 

We  have  already  observed  in  Art.  9,  Chap.  I.,  that  we  can 
always  find  the  strains  in  the  pieces  which  meet  at  an  apex, 
provided  only  two  are  unknown.  Now  in  the  strain  diagram 
to  Fig.  IY.,  we  readily  determine  the  strains  in  X  «,  Y  a,  X  b, 
a  b,  Y  c  and  b  c  successively,  and  arrive  finally  at  apex  2,  where 
we  have  the  two  known  strains  in  X  5  and  b  £,  and  wish  to  find 
the  strains  in  three  pieces,  viz.,  X  cZ,  d  Ji  and  c  h.  At  first  sight 
this  seems  impossible.  If,  however,  we  assume  that  the  pieces 
of  the  frame  can  take  only  strains  of  a  certain  kind,  as,  for 
instance,  hd  only  tension,  and  not  compression,  the  problem  is 
perfectly  determinate.  This  assumption  is  easily  realized  in 
practice.  Thus  if  h  d  is  a  rod  of  small  diameter,  it  cannot 
act  as  a  compression  member  at  all.  Moreover,  the  strain 
of  tension  in  h  d  must  evidently  be  precisely  equal  to  that  in  1 1 
b  c,  already  found.  We  have  then  to  form  a  closed  polygon  u 
with  the  weight  at  2  and  the  known  strains  in  X  b  and  b  c,  whose  * 
other  three  sides  shall  be  parallel  to  X  d,  h  d  and  c  h  respec- 
tively, and  in  which,  moreover,  the  strain  in  h  d  shall  be  equal 
to  that  in  b  c,  and  where  both  these  strains  must  be,  when  the 
polygon  is  followed  round  according  to  rule,  tensile.  We  have 
evidently,  then,  in  accordance  with  these  conditions,  only  tlio 
polygon  2  X  d  h  c  b  X,  thus  finding  the  point  d,  from  which 
we  can  now  proceed  to  find  e,  etc.  The  points  a,  b,  d  and  e  are 
evidently  in  the  same  straight  line  parallel  to  c  h.  This  point 
is  one  of  importance,  and  the  reader  should  carefully  follow 
the  above  remarks  with  the  aid  of  the  Fig. 

The  strain  diagram  thus  constructed  shows  us  many  facts 


350  NOTE  TO  CHAP.  I.  [APPENDIX. 

about  the  system  not  otherwise  apparent.  Tims  ~b  c,  c  h  and  h  d 
are  in  equilibrium  with  the  load  at  2.  Again,  a  b  and  b  c  are 
in  equilibrium  with  Y  a  minus  Y  <?,  as  also  are  hd  and  de  with 
Jc  e  minus  kh.  Also  Teh,  h  c,  Y  c  and  Y  Jc  are  in  equilibrium, 
and  Yc,  cb  and  X  b  are  in  equilibrium  with  the  reaction  minus 
the  weight  at  1,  or  with  the  shear  to  the  right  of  1.  This  last 
principle  is  general.  When  a  section  can  ~be  made  entirely 
through  a  structure,  the  strains  in  the  pieces  cut  are  in  equi- 
librium with  the  shear  at  the  section.  If  only  three  pieces  are 
cut,  then,  by  taking  as  a  centre  of  moments  the  point  of  inter- 
section of  any  two,  we  can  easily  find,  knowing  the  moment  of 
the  shear,  the  strain  in  the  third. 

Thus  we  have  the  general  and  easy  method  of  calculation 
given  in  Art.  14,  Chap.  I.  The  moment  of  the  shear  is,  of 
course,  the  sum  of  the  moments  of  all  the  exterior  forces  be- 
tween the  section  and  one  end. 

We  have  then  two  methods,  one  graphic  and  one  by  calcula- 
tion, by  which  we  can  find  the  strains  in  every  kind  of  simple 
truss  which  can  ever  occur  in  practice.  By li  simple  "  we  mean 
merely  resting  at  the  supports,  or  not  acted  upon  at  the  ends  by 
a  couple  or  moment,  as  is  the  case,  for  instance,  in  the  continuous 
girder. 

When  the  structure  is  unsymmetrical,  or  complex,  the  deter- 
mination of  the  different  lever  arms  is  often  very  tedious,  involv- 
ing a  good  deal  of  trigonometrical  computation.  On  the  other 
hand,  the  frame  can  always  from  its  known  proportions  be 
easily  and  accurately  drawn  to  scale,  and  then  the  exterior 
forces,  whatever  their  relative  intensity  or  directions,  can  be 
laid  off,  and  the  strains  at  once  determined.  Here  we  see,  then, 
one  of  the  great  advantages  of  our  graphical  method.  An 
unsymmetrical  frame  and  different  directions  of  the  forces 
requires  no  more  time  or  labor  than  a  more  simple  case. 

7.  Application  to  Bridges— Bow-string  Girder. — In  Art. 
12,  Chap.  I.,  we  have  alluded  to  this  application,  and  shown 
how  by  two  strain  diagrams  only  we  can  completely  calculate  a 
bridge  of  any  length.  As  this  application  is  so  important,  and 
as  the  method  is  stated  by  several  authors  to  be  inapplicable  to 
bridges,*  or,  at  best,  to  be  unsatisfactory,  we  will  here  call  more 

*  Iron  Bridges  and  T&w/s— Unwin — p.  143.  Economics  of  Construction — 
Bow-  -p.  61. 


APPENDIX.] 


NOTE   TO    CHAP.  I. 


351 


special  attention  to  the  points  to  be  observed  in  the  tabulation 
of  the  strains.  There  is,  indeed,  no  more  satisfactory,  complete 
and  rapid  method  for  the  solution  of  bridge  girders  generally 
than  that  afforded  by  the  graphic  method. 

As  an  example,  let  us  take  the  Bow-string  Girder  given  by 
Stoney,  p.  131.  Span,  80  ft.,  divided  into  8  panels ;  rise  of 
bow,  10  ft.  Load,  10  tons  at  each  lower  apex. 

We  construct  the  two  strain  diagrams*  given  in  Fig.  V., 
PL  2,  viz.,  one  for  the  load  P7  at  the  first  apex,  and  one  for  the 
load  at  the  last  apex,  P,.  Referring,  if  necessary,  to  Art.  12, 
Chap.  I.,  the  reader  can  easily  follow  out  these  diagrams.  We 
then  scale  off  the  strains,  and  obtain,  for  the  strains  in  the 
diagonals — 


ab 

be 

cd 

de 

ef 

fff 

ffh 

FT 

-2.7 

-11.4 

+  4.8 

-  4.3 

+  2.4 

-2.3 

+  1.4 

Pi 

-0.4 

+  0.23 

-0.56 

+  0.51 

-0.9 

+  0.88 

-  1.4 

Now  from  the  strains  thus  obtained  for  these  two  weights 
we  can  easily  obtain  all  the  others. 

Thus,  as  the  end  reactions  are  inversely  as  the  distances  of 
the  weight  from  the  ends,  the  reaction  at  the  left  end  due  to 
P2  will  be  twice  that  due  to  Pa.  For  Ps  three  times  that  due 
to  P,.  The  strains  will  therefore  be  twice  and  three  times 
those  due  to  P1?  until  we  arrive  at  the  weights  P1  and  Pa  re- 
spectively. So  also  for  P6  the  reaction  at  the  right  is  twice 
that  due  to  P7,  and  the  strains  are  therefore  double  up  to  the 
weight  P6.  To  the  right,  then,  of  P6  the  strains  are  twice 
those  due  to  P7,  and  to  the  left  of  P6  they  are  six  times  those 
due  to  P!.  Take,  for  instance,  P5.  The  right  reaction  is  f ths 
of  the  apex  load,  and  the  right  reaction  of  P7  is  -|th  of  that 
load.  For  P5,  then,  the  strains  in  all  pieces  to  the  right  of  that 
weight  are  3  times  those  due  to  P7.  Again,  the  left  reaction 
is  for  P5  |ths  the  apex  load.  But  the  left  reaction  for  Pl 
is  -Jth  the  same  load.  The  strains  then  in  all  the  pieces  to  the 
left  of  P5  are  5  times  those  due  to  P!.  So  for  any  other  load. 
We  can  therefore  form  at  once  the  following  table : 

*  Strain  diagrams  in  Fig.  V.,  and  also  in  Fig.  VL,  are,  for  obvious  reasona, 
drawn  to  different  scales. 


352 


NOTE   TO    CHAP.  I. 


[APPENDIX. 


« 

PI 

P2 

P3 

P4 

£5 

P6 

P7 

Uniform 
Load. 

Max. 
Comp. 
+ 

Max. 
Tens. 

Total 
Strains. 

ab 

-0.4 

-  0.8 

-  1.2   -  1.6 

-2.0 

-2.3 

-2.7 

-  8.25 

1-n.o 

-  19.25 

be 
cd 

+0.23 
^.56 

+  C.5 
—  1.1 

+  0.7 
—  1.7 

+  0.9 
-  2.2 
+  2.0 

+  1.1 

+  1.4 

-11.4 
+4.8 

-4.9 

+  4.8   -  11.4 

+  4.bj  -  11.8 

-  16.3 

-  2.8 

-3.4 

-5.2 

-  17.0 

de 
ef 

+0.51 
^0^ 

+  1.0 
-  1.8 

+  1.5 

+  2.6-8.6 

-4.3 
+  2.4 

-3.97 

-4.8 

+  7.6 
+  7.1 

-  12.9 

-  16.9 
-  18.3 

-  2.7   -  3.6 

-4.5 

+  4.7 

-  13.5 

fa\ 

d 

+0.88 

+  1.8 

+  2.6  +  3.5 

-6.9 

-4.6 

-2.3 

—  0.55 
-  4.2 

+  8.8 

-  13.8 

-20.5 

-1.4 

-2.8 

-4.2 

-5.6 

+  ,2 

+  ,8 

+1.4 

+  8.4 

-14.0 

-  18.2 

In  the  columns  for  Pt  and  P7  we  put  the  strains  already 
found  by  diagram.  The  strains  for  P2  on  the  entire  left  half 
will  be  double  those  for  Px ;  for  P3  three  times,  and  for  P4  four 
times  those  for  Px.  We  have  therefore  at  once  the  columns 
for  P1?  P2,  P3  and  P4.  Now  for  P5  we  see  from  the  Fig.  that 
the  strains  in  diagonals  ef  and  fg  must  both  be  tension. 
From  the  left  end,  then,  as  far  as  ef,  the  strains  are  4  times 
those  due  to  Pl5  and  from  the  right,  as  far  asfg,  3  times  those 
due  to  P7.  We  thus  obtain  the  column  for  P5.  In  the  same 
way  for  P6,  all  above  or  to  left  of  cd  are  6  times  Pl5  all  below 
or  to  right  of  de  twice  P7.  Thus  we  fill  out  the  whole  table. 
Adding  now  all  the  tensions  and  compressions  in  each  piece,  we 
obtain  the  maximum  strains  of  each  kind  due  to  the  live  load, 
as  given  in  the  last  two  columns  but  one.  Suppose  now  the 
dead  load  or  weight  of  the  girder  itself  to  be  fths  of  the  roll- 
ing or  live  load.  We  have  only  to  take,  then,  f  ths  the  sum  of 
these  last  two  columns  and  we  have  the  strains  due  to  uniform 
or  dead  load,  as  given  in  the  fourth  column  from  the  right. 

We  can  now  easily  obtain  the  total  strains.  Thus  the  ten- 
sion in  a  1)  due  to  the  live  load  only  is  11  tons.  The  tension 
due  to  the  dead  load  is  8.25  tons.  Total  greatest  strain  which 
can  ever  come  upon  a  5,  then,  is  19.25  tons  tension.  No  com- 
pression can  ever  come  on  this  |)iece ;  it  does  not  need,  there- 
fore, to  be  counterbraced.  On  the  other  hand,  all  the  other 
diagonals,  except  perhaps  cd,  must  be  counterbraced,  as  the 
maximum  compression  due  to  the  live  load  overbalances  the 
constant  tension  of  the  dead.  Had  the  dead  load  been  taken 
much  greater  than  the  live,  the  diagonals  might  always  have 
been  in  tension.  Hence  the  appropriateness  of  this  class  of 
girder  for  long  spans. 


APPENDIX.] 


NOTE  TO  CHAP.  I. 


353 


We  see  also  from  the  table  just  what  weights,  and  where 
placed,  give  the  greatest  strain  of  eacli  kind-  in  any  piece. 

§.  Strains  in  tlie  Flanges. — The  method  is  precisely  simi- 
lar for  the  flanges.  Thus  we  scale  off  from  our  diagrams — 


Xa 

Xd 

JLd 

*/ 

Th 

P! 

+  2.82 

+  3.08 

+  3.47 

+  4.11 

+  5.11 

PT 

+  19.7 

+  21.6 

+  10.4 

+6.8 

+  5.1 

Ya 

Yc 

Ye 

?9 

Pi 

-  2.52 

-3.01 

-3.62 

-  4.46 

P7 

-  17.6 

-  13.1 

-7.9 

-5.7 

Tabulating  these,  we  obtain  the  following  table : 


Flanges. 

PI 

! 

P2 

P3 

P4 

P5 

P6 

P7 

Uniform 
live 
load. 

Uniform 
dead 
load. 

Total 

Strains. 

Xa 

+2.82 

+  5.6 

+  8.5 

+11.3 

+  14.1 

+  16.9 

+19.7 

+  78.9 

+  59.1 

+  138 

X6 

+3,08 

+  6.2 

+  9.2 

+12  3 

+  15.4 

+18.5 

+21.6 

+  86.3 

+  64.7 

+  151 

Xrf 

+3.47 

+  6.1) 

+  10.4 

+13.9 

+  17.3 

+20.8 

+10.4 
1^.8 
+5.1 

+  83.2 

+  62.4     +  145.6 

x/ 

~XA 

+4.11 

+  8.2 

+  12.3 

+16.4J+20.5 

+  13.7 
+  10.2 

+  82.0 
"+81.6 

+  61.5 
+  61.2 

+  143.5 

+5.11 

+  10.2 

+  15.3 

+20.4 

+15.3 

+  142.8 

Ya 

-2.52 
-3.01 

-5.0, 

-7.G 

-9.0 

-10.1 
-12.0 

-12.6 

-15.0 

-15.1 

^-IsT 

-17.6 

-70.5 
-76.2 

-52.8 

-128.3 

Yc 

-6.0, 

-13.1 

-57.1 

-  133.3 

Ye 
Y7 

-3.62 
^A6 

-7.  2,  -10.9 

-14.5 

-18.1 

-15.9 

-7.9 

-78.1 

-58.5 

-  136.6 

-8.9  -13.4 

-17.8 

-17.1 

-11.4 

-5.7 

-  78.8 

-59.1 

—  137.9 

This  table  is  obtained  precisely  as  before.  Thus  for  P6  the 
strains  in  Xa  and  Xb  are  multiples  of  P1?  while  those  in  the 
other  flanges  are  multiples  of  P7.  So  also  for  Y  c  and  Y  e.  We 
f  see  at  once  that  the  greatest  strains  are  for  full  load,  since  for 
all  loads  the  upper  flanges  are  always  compressed  and  the 
lower  extended.*  The  above  is  sufficient  to  illustrate  fully  the 
application  of  our  method  to  bridges.  It  is  evidently  appli- 

*  A  more  convenient  form  of  tabulation  is  to  put  the  weights  in  the  left 
vertical  column  and  the  pieces  in  the  top  horizontal  line.     The  numbers  can 
then  be  more  easily  added. 
23 


354  NOTE   TO    CHAP.  I.  [APPENDIX. 

cable  to  any  structure  where  the  reactions  are  inversely  as  the 
distances  from  the  end.  The  strains  due  to  the  first  and  last 
weights  are  all  that  we  need  in  order  to  thoroughly  solve  any 
case  of  the  kind.  It  is  advisable,  however,  to  construct  a 
third  diagram  for  an  intermediate  weight,  in  order  to  serve  as 
a  check  upon  the  others.* 

9.  Method  of  Calculation  by  Moments. — AVe  may  illus- 
trate here  the  method  of  calculation  by  moments  referred  to  in 
Art.  14,  Chap.  L,  a  little  more  fully.  Thus  in  the  example 
above,  Fig.  V.,  suppose  we  wish  the  strains  due  to  P{.  Reac- 
tion at  left  end  is  evidently  £th  of  10  tons  =  1.25  tons.  Con- 
ceive the  lower  flange  Y  a  cut.  Rotation  would  evidently  take 
place  about  apex  a,  and  we  have,  therefore,  strain  in  Y  a  x  its 
lever  arm  from  apex  a  =  1.25  x  5.  The  depth  of  truss,  or 
lever  arm  of  Y$,  from  apex  a,  is  2.58  feet.  Hence  we  have 

'  ,  K0 —  =  strain  in  Y  a  —  2.42  tons. 
a.Do 

This  strain  is  evidently,  by  reason  of  the  direction  in  which 
the  two  portions  of  the  truss  would  rotate  about  a,  tension.  In 
like  manner,  for  upper  flange  X  &,  if  we  know  the  lever  arm  of 
this  flange  from  the  opposite  apex,  we  can  easily  find  the  strain  ; 
for  the  diagram  shows  that  X  5,  b  c  and  Y  c  are  in  equilibrium 
with  the  reaction,  and  hence,  if  we  take  the  point  of  moments 
at  the  intersection  of  the  two  pieces  b  c  and  Y  c,  the  moments 
of  these  pieces  are  zero,  and  we  have  remaining  only  the  mo- 
ment of  the  strain  in  X  b  balanced  by  the  moment  of  the  reac- 
tion. 

Again,  if  X  b  and  Y<?  are  thus  found,  and  if  these  two  strains, 
together  with  the  reaction,  are  in  equilibrium  with  the  diago- 
nal b  c,  we  can  find  the  strain  in  this  diagonal  by  taking  the 
apex  d  as  a  centre  of  moments.  The  moment  of  Y  c  then  is 


*  It  may  also  be  well  to  notice  here  that  the  practice  of  deducing  in  the 
tabulation  the  dead  load  from  the  live  load  strains  is  not  strictly  accurate,  as 
the  live  load  acts  at  the  lower  apices  only  [or  at  the  upper  apices  only,  if  the 
bridge  is  under  grade],,  while  the  dead  load  is  distributed  along  both  flanges, 
and  acts  at  both  upper  and  lower  apices. 

In  every  case,  however,  the  greater  portion  of  the  dead  load,  say,  for  in- 
stance, fds  of  the  whole,  owing  to  the  track,  platform,  cross-girders,  etc., 
acts  at  the  same  apices  as  the  live  load  itself ;  and  the  error  is  in  any  case 
very  slight,  and  practically  of  no  account. 


APPENDIX.]  NOTE   TO    ClfrAP.  I.  355 

zero,  and  we  have  the  moment  of  the  strain  in  ~b  c  balanced 
by  the  moment  of  the  reaction  and  the  moment  of  X  5  ;  the 
first  causing  compression,  the  second  tension,  and  the  difference 
then  giving  the  resultant  moment  strain,  which,  divided  by  the 
lever  arm  of  &  <?,  gives  the  strain  itself. 

The  method  is  easy  of  application,  but,  as  we  have  already 
remarked,  the  determination  of  the  lever  arms  for  each  case  is 
frequently  tedious.  "  These  may,  however,  be  scaled  off  from 
the  frame  diagram  with  sufficient  accuracy  in  practice. 

As  before,  we  need  only  the  strains  due  to  the  first  and  last 
weight,  and  can  then  form  our  tabulation  as  above.  This 
tabulation  we  can  also  check  by  finding  the  strains  due  to  uni- 
form load  independently,  and  seeing  whether  it  agrees  with 
the  sum  of  the  separate  apex  weight  strains. 

Thus,  for  all  the  weights  acting,  suppose  we  wish  the  strain 
in  Y  g.  The  lever  arm  of  Y  g  is  9.85  feet.  We  have  then  re- 
action =  35  tons,  multiplied  by  35  feet  =  1225.  This  must  be 
diminished  by  P7  X  25  =  250,  P6  x  15  =  150,  and  P5  x  5  = 
50.  We  have  then  1225  -  450  =  775,  which,  divided  by  9.85, 
gives  78.7  tons  tension  in  Y  </,  agreeing  with  our  tabulation 
above. 

For  the  method  of  calculation  by  resolution  of  forces,  see 
Art.  16  of  this  Appendix. 

10.  Girder  with  Straight  Flanges. — In  such  a  case,  as  the 
lever  arms  are  at  once  known  and  are  constant,  the  above 
method  is  of  very  easy  application.  In  this  case  the  strains  in 
the  diagonals  are  best  found  by  multiplying  the  shear  by  the 
secant  of  the  inclination  of  the  diagonal  with  the  vertical.* 
Thus,  if  this  angle  is  45°,  we  have  simply  to  multiply  the  shear 
at  any  point  by  1.4142,  and  we  have  at  once  the  strain  in  the 

*  This  is  but  a  particular  result  of  the  general  method  of  moments.  Thus, 
for  any  diagonal,  as  ab  (Fig.  VII.),  according  to  our  rule,  we  take  the  centre  of 
moments  at  the  intersection  of  the  two  other  sides  cut  by  a  section  through 
the  truss,  viz. ,  the  flanges.  But  these  two  sides  are  here  parallel,  hence  their 
intersection  is  at  an  infinite  distance.  The  lever  arm  of  a  b  is  then  GO  x  cos  </-, 
6  being  the  angle  with  the  vertical.  If  the  weight  PI  acts,  we  have  then, 
calling  the  reaction  R,  R  x  GO  —  P  GO  =  S  cos  </>  x  GO,  where  S  is  the  strain  in 

a  b.     This  can  be  put  (R  —  P)  GO  =  S  cos  0  x  co,  hence  S  =  - .      But 

GO  1 

R  —  P  is  the  shear  at  5,  C08  >  x  00=       T  =  sec  0  ;  hence  we  have  only  to 

multiply  the  shear  by  the  secant. 


356  NOTE  -TO  ciiAr.  i.  [APPENDIX. 

diagonal  at  this  point.  The  shear  is  always  in  such  cases  the 
reaction  at  the  end  minus  the  weights  between  that  end  and 
the  apex  in  question. 

The  flanges  are  easily  obtained  by  moments,  as  above. 

The  following  points  need  attention,  however.  First,  if 
there  are  two  or  more  systems  of  diagonals,  as  represented  in 
PL  2,  Fig.  YIL,  by  the  full  and  dotted  diagonals  (omitting  the 
upright  lines),  we  must  find  the  strains  for  each  system  sepa- 
rately, and  then  add  them  together.  Thus,  if  the  strains  found 
in  a  c  and  c  e*  etc.,  for  one  system,  are  50  and  60  tons,  and 
those  in  df  and  fg,  for  the  other,  are  40  and  TO  tons,  we  have, 
when  the  two  systems  are  combined,  d  c  =  a  c  +  df  =  50  +  40 
=  90,  cf=  df+ce  =  ±0  +  60  =  100,  fe  =  ce  +.fg  = 
60  +  70  =  130,  and  so  on.  This  holds  true,  of  course,  whether 
the  strains  are  obtained  by  calculation  or  diagram.  Thus,  for 
a  lattice  girder,  we  calculate  or  diagram  each  system  ~by  itself, 
and  then  the  strain  in  any  flange,  when  the  two  are  combined, 
is  equal  to  the  sum  of  the  strains  on  that  flange  due  to  each 
system  of  triaiigulation  which  includes  it. 

There  is  another  point  to  be  observed  in  connection  with 
the  system  known  as  the  Howe  or  Pratt  Truss.  Inserting  the 
dotted  verticals  into  our  Fig.,  we  have  this  system  of  square  pan- 
elling. Let  us  suppose  that  the  diagonals  take  tension  only, 
and  the  verticals  compression  only. 

Now  for  a  weight  at  apex  9  of  10  tons,  we  have  a  right  re- 
action of  1  ton,  which,  running  through  the  system,  causes  strain 
in  the  diagonal  of  f  P4.  For  the  flange  D,  then,  our  point  of  mo- 
ments is  at/,  and  if  the  height  of  truss  is  equal  to  panel  length, 

1  x  50 
viz.,  10  feet,  we  have  the  strain  in  D  =  — — —  =  5   tons,   for 

P9.  In  the  same  way  for  P8,  we  have  for  D  10  tons  ;  for  P7, 15 
tons ;  for  P6,  20  tons ;  for  P5,  25  tons.  For  P4,  on  the  other 
hand,  we  have  a  left  reaction  of  4  tons,  which  causes  strain  in 
diagonal  e  &,  and  for  this  weight  and  all  succeeding  weights 
our  point  of  moments  for  D  is  then  at  e.  We  have  then  P4 

4  x  —  =  24  tons ;  for  P3,  18  tons ;  for  P2, 12  tons ;  and  for 

P!,  6  tons. 

For  all  these  weights,  then,  acting  together,  we  have  135  tons 
strain  in  D. 


APPENDIX.]  NOTE   TO    CHAP.   I.  .357 

But  for  all  tlie  weights  acting  together,  it  is  evident  that 
only  all  the  braces  sloping  each  way  from  the  centre  are 
strained.  Hence  e  Tc  is  not  strained,  and  our  point  of  moments 
is  for  D  always  at/!  Thus  for  total  load  we  have  strain  in 

45  x  50  -  10  x  40  -  10  x  30  - 10  x  20  -  10  x  10  ' 
D  =  -  1Q  -  =  125  tons, 

whereas  we  found  by  addition  of  the  several  weights  135  tons. 

There  is  thus  an  ambiguity  in  this  class  of  bracing  as  to  the 
way  in  which  the  strains  may  go.  Two  symmetrical  weights, 
as  9  and  1,  may  either  go  left  and  right  directly  to  the  abut- 
ments or  a  portion  of  each  go  towards  the  centre.  The  inter- 
mediate diagonals  may  be  either  all  strained  or  not  strained  at 
all.  The  strains  may  go  partly  in  one  way  or  partly  in  the 
other.  We  should  then  not  rely  on  our  summation  of  the  sepa- 
rate weights,  but  always  check  them  by  calculation  or  diagram 
for  the  total  load  also,  and  take  the  greatest  strain*.  Practi- 
cally, for  long  spans,  it  is  very  rare  that  the  difference  is  of  any 
importance. 

In  diagraming  by  our  method  such  a  system  of  bracing  as 
the  above,  we  should  consider  but  one  series  of  braces,  viz., 
those  strained  by  the  uniform  load  alone.  Thus,  for  our  Fig. 
and  loads  on  the  lower  apices,  we  should  take  only  the  diago- 
nals parallel  to  fh  on  the  left  of  centre,  and  yP4  on  the  right. 
If,  on  the  other  hand,  the  verticals  are  ties  and  the  diagonals 
struts,  we  should  retain  only  those  parallel  to  c  &  on  the  left, 
and  those  parallel  to  k  e  on  the  right  of  centre.  The  others 
are  to  be  omitted.  Then,  the  tabulation  being  formed,  if  in 
any  diagonal  a  strain  may  occur  of  reverse  character  to  that 
which  it  is  intended  to  resist,  a  counterbrace  must  be  inserted 
in  this  panel  to  take  this  reverse  strain. 

As  in  our  examples  we  have  taken  always  a  triangular  system 
of  bracing,  it  is  important  that  the  reader  clearly  understand 
the  method  to  be  pursued  in  other  forms.  For  the  rectangular 
system  of  bracing  generally,  the  point  where  for  uniform  load 
the  shear  is  zero  is  the  point  from  which  the  braces  must  slope 
both  ways.  The  other  diagonals,  or  the  count  erbr  aces,  are  then 
omitted  in  both  calculation  and  diagram,  and  replaced  from  the 
tabulation  when  necessary  to  replace  a  strain  of  the  reverse 
character  to  that  which  the  braces  are  intended  to  sustain. 

Attention  to  the  above  points  will  enable  us  to  both  calculate 


353 


NOTE   TO   CHAP.  I. 


[APPENDIX. 


and  diagram  with  ease  and  accuracy  any  form  of  truss  which 
occurs  in  engineering  practice. 

11.  In  the  bow-string  girder  represented  ir  Fig,  Y.  it  is 
evident  that  the  bottom  flange  serves  merely  to  resist  the  thrust 
of  the  bow  and  keep  it  from  spreading.  It  adds  nothing  to  the 
supporting  power  of  the  combination.  We  might  remove  it 
entirely  and  replace  it  by  abutments  which  would  equally  well 
sustain  this  thrust,  and  if  we  then  introduced  a  horizontal  flange 
at  crown,  and  inserted  diagonals  between  for  stiffness,  we  should 
have  the  form  of  braced  arch  given  in  Chap.  I.,  Fig.  5  (c).  If, 
however,  we  should  resist  the  thrust  of  the  bow  by  an  inverted 
arc,  it  would  answer  the  same  purpose  as  the  bottom  flange, 
and  we  should,  in  addition,  double  the  supporting  power. 

We  have  illustrated  this  in  Fig.  YI. 

The  span  is  the  same  as  before.  The  lower  apices  only  are 
supposed  to  be  loaded,  for  comparison.  [Properly,  we  should 
have  distributed  the  load  over  both  upper  and  lower  apices.] 
The  rise  of  each  arc  is  one-half  as  great  as  before,  or  5  f  t%  only, 
thus  making  the  total  depth  the  same  as  in  the  preceding 
case. 

By  means  of  two  strain  diagrams,  we  find  the  strains  due  to 
PiandP7.  Thus: 


xa 

Xb 

Xd 

x/ 

Xh 

Ya 

Y  c 

Y  e 

Y<7 

Pi 

+2.3 

+2.6 

+3.2 

+3.7 

+5.0 

-2.25 

-2.87 

-3.48 

-4.32 

P7 

+16.1 

+17.8 

+9.1 

+5.7 

+4.5 

-15.75 

-12.35 

-7.3 

-4.8 

Then,  precisely  as  in  the  preceding  Art.,  we  can  fill  out  our 
table  of  strains.  This  the  reader  can  now  easily  do  for  himself. 
We  thus  find,  for  a  uniform  dead  load  f ths  the  live  load,  the 
total  maximum  strains  below. 


Xa 

Xb 

Xd 

x/ 

Xh 

Y  a 

Yc 

Y<5 

Y£ 

+112.7 

+126.7 

+133.5 

+127. 

+134.7 

-110.1 

-126.7 

-129.3 

-125.6 

Comparing  these  with  the  corresponding  strains  for  the  bow- 


APPENDIX.]  NOTE    TO    CHAP.  I.  359 

string,  we  find  that  they  are  very  much  less  in  every  piece.  In 
fact,  there  is  a  total  gain  of  over  10  per  cent.,  and  that,  too,  not- 
withstanding that  the  rise  of  each  arc  is  only  half  that  in  the 
first  case.  Had  we  taken  a  double  depth,  the  saving  would 
have  been  very  great,  and  as  in  this  case  also,  for  a  long  span 
and  relatively  large  dead  load,  the  diagonals  would  always  be  in 
tension,  the  increased  length  of  these  last  would  be  no  dis- 
advantage. 

12.  The  above  construction  is  worthy  of  the  careful  consid- 
eration of  the  bridge  builder.  It  peculiarly  recommends  itself 
for  long  spans,  and  has  several  important  advantages  possessed 
by  no  other  form  of  truss.  For  long  spans  the  strains  in  the 
flanges  are  nearly  uniform.  The  diagonals  are  less  strained 
than  in  any  other  system,  and  are  always  in  tension.  Every 
member  acts  to  support,  as  well  as  to  strengthen.  The  height* 
is  everywhere  proportional  to  the  maximum  moment  of  the  ex- 
terior forces.  The  load  is  distributed  along  the  neutral  axis, 
thus  securing  the  maximum  of  rigidity  ;  while  the  neutral  axis 
itself  passes  through  the  points  of  support. 

This  construction  is  known  in  Germany,  from  the  name  of  its 
inventor,  as  Paulas  Truss.  Upon  this  system  are  the  double 
track  bridge  over  the  Isar  at  Grossheselohe,  2  spans  of  170.6 
ft. ;  a  large  number  of  smaller  bridges,  such  as  one  over  the 
Rodach,  109  ft.  span  ;  over  the  Main  in  Schweinfurt,  116.4ft. 
span ;  and  especially  one  over  the  Rhine  at  Mayence,  of  32 
spans,  4  of  345  ft.,  6  of  116  ft.,  20  of  50  ft.,  and  2  of  82  ft. ;  all 
upon  the  same  system. 

In  England,  we  might  notice  the  famous  bridge  over  the 
Tamar  at  Saltash,  near  Plymouth,  whose  two  principal  spans 
are  455  ft.,  which  is  also  constructed  upon  this  system. 
Finally,  we  may  mention  the  bridge  over  the  Elbe,  near 
Hamburg,  the  three  principal  spans  of  which  are  325  ft.  each. 

In  this  latter  structure  both  the  upper  and  lower  members 
are  braced  or  ribbed  arches,  of 'a  constant  depth  of  about  10  ft., 
a  combination  which,  for  long  spans,  seems  most  excellent.  A 
single  arch  alone,  similar,  for  example,  to  the  steel  arch  over  the 
Mississippi,  by  Capt.  Eads,  would  have  required  heavy  abut- 
ments. 

The  same  arch  inverted  would  have  required  equally  heavy 
anchorages.  The  combination  does  away  with  both.  The 


360  NOTE  TO  CHAP.  I.  [APPENDIX. 

thrust  of  the  upright  arch  is  opposed  by  the  pull  of  the  inverted 
one,  all  the  advantages  of  Pauli's  system  are  obtained,  and 
there  are  no  temperature  strains  such  as  occur  in  the  single 
arch,  while  the  bracing  is  reduced  to  a  minimum.  At  the  same 
time  all  the  rigidity  due  to  the  arch  is  obtained.* 

13.  In  the  construction  of  the  diagrams,  care  should  be  exer- 
cised in  the  selection  of  the  scales,  that  the  frame  diagram  may 
be  large  enough  to  secure  the  desired  accuracy.  Lines  should 
be  drawn  very  fine  with  a  hard,  sharp-pointed  pencil,  so  as  to  be 
scarcely  discernable,  and  their  intersections  accurately  marked 
by  needle  point. 

With  an  accurate  scale  and  good  instruments,  strains  can  be 
taken  off  in  nearly  every  practical  case  to  hundredths  of  a  ton 


*  Compare  Long  and  Short  Span  Railway  Bridges,  by  John  A.  Roebling,  C.E. 
— In  this  work,  Mr.  Roebling  proposes  a  system  in  principle  essentially  the 
same  as  the  above,  to  which  he  gives  the  name  of  "Parabolic  Truss."  He, 
however,  constructs  the  arch  of  channel  irons  bolted  to  the  sides  of  a  straight 
truss,  the  sole  office  of  which  is  to  give  rigidity  to  the  system.  Also,  claiming 
that  iron  in  the  shape  of  wire  will  safely  sustain  three  times  as  much  as  in  the 
shape .  of  bars  or  rods,  he  introduces  a  wire  cable  in  place  of  the  inverted 
braced  arch. 

It  will  thus  be  seen  that  for  rigidity  the  system  is  wholly  dependent  upon 
extraneous  members,  such  as  the  auxiliary  truss  and  the  tower  stays,  which  are 
liberally  introduced.  By  dividing  the  material  composing  the  upright  arch 
into  two  portions,  bracing  between  them,  and  thus  forming  a  braced  arch  sim- 
ilar to  Capt.  Eads,  the  stays  and  stiffening  truss  might  be  entirely  dispensed 
with,  the  construction  greatly  simplified  in  the  number  of  its  members,  and 
the  bracing  reduced  to  a  minimum.  If,  also,  as  claimed  by  Capt.  Eads,  the 
conditions  for  cast  steel  are  just  the  reverse  of  iron,  and  it  is  most  advantageous 
to  use  it  in  compression,  then  it  seems  that  such  a  modification  of  Mr.  Roeb- 
ling's  design  with  wire  cable  and  a  cast-steel  braced  arch  would  better  sustain 
the  thesis  with  which  his  work,  above  quoted,  opens,  viz.  :  that  "the  greatest 
economy  in  bridging  is  only  to  be  obtained  by  a  judicious  application  of  the  Para- 
bolic Truss." 

Such  a  combination  of  the  suspension  and  upright  arch  would  seem  to  avoid 
the  principal  objections  urged  against  each  separately. '  The  anchorages  and 
abutments  are  dispensed  with,  the  greatest  rigidity  is  secured  with  the  mini- 
mum of  bracing,  and  the  material  is  used  in  the  most  advantageous  way.  In 
addition  to  the  advantages  of  Paulas  system-being  secured,  we  have  the  ease 
of  erection  of  the  suspension  system  combined  with  the  rigidity  of  the  arch. 
The  system  is  self -balancing,  and  practically  unaffected  by  changes  of  temper- 
ature. 

For  the  practical  details  of  construction  of  such  a  system,  the  reader  can 
with  profit  consult  Mr.  Roebling' s  work,  above  quoted.  They  will  be  found 
to  be  neither  expensive  nor  difficult  of  execution. 


APPENDIX.]  NOTE   TO    CHAP.  I.  361 

with  perfect  accuracy.  The  use  of  parallel  rulers  is  not  to  be 
recommended.  The  T  square,  triangle  and  drawing-board  are 
far  preferable.  It  should  be  remembered,  finally,  that  careful 
habits  of  manipulation,  while  they  give  constantly  increased 
skill  and  more  accurate  results,  affect  in  no  degree  the  rapidity 
and  ease  with  which  those  results  are  obtained. 


362  NOTE   TO   CHAP.  H.  [APPENDIX. 


NOTE  TO  CHAPTER  II. 

14.  The  reader  will  observe  that  in  Chapter  I.  we  had  given 
forces  acting  at  certain  points  of  a  given  frame,  and  we  found 
by  simple  resolution  of  forces  the  strains  in  the  pieces  of  that 
frame.  In  Chapter  II.  we  have  given  forces  acting  in  certain 
directions,  and  having  assumed  the  strains,  we  find  the  equi- 
librium, polygon  or  frame,  which,  having  its  angles  on  these 
force  directions,  and  having  these  strains,  will  hold  the  given 
forces  in  equilibrium.  Thus  in  Figs.  12  (b)  and  (c),  PL  III.,  of 
the  text,  by  choosing  a  pole  and  drawing  lines  to  the  forces  in 
the  force  polygon  (a),  we  virtually  assume  the  strains  which 
are  to  act  upon  our  frame.  Then  lines  parallel  to  these  strains 
in  (b),  forming  a  polygon  whose  angles  are  upon  the  forces, 
must  give  us  the  frame  which  holds  these  forces  in  equilibrium, 
provided  we  close  the  polygon  by  a  line  and  apply  at  the  ends 
forces  which  balance  each  other  horizontally,  and  whose  com- 
ponents parallel  to  the  resultant  of  the  forces  balance  the 
forces. 

Thus  the  polygon  maficdenm  is  a  frame  along  whose 
sides  the  forces  S0  Sl5  etc.,  act,  and  whose  reactions  at  the  sup- 
ports m  and  n  must  then  be  a  o  and  5  a,  as  given  in  (a). 

This  frame — keeping  the  same  pole,  that  is,  the  same  strains 
— we  may  put  anywhere  in  the  plane,  its  angles  being  always  on 
the  forces,  and  its  sides  always  respectively  parallel,  though 
varying  in  length  according  to  the  position  assumed. 

We  might  also  have  assumed  different  strains,  that  is,  taken 
a  different  pole,  and  constructed  a  different  frame;  but  evi- 
dently the  end  reactions  will  not  be  altered,  and  will  be  always 
equal  to  a  0  and  5  a,  as  given  in  (a). 

The  peculiarities  of  the  frame  thus  obtained  are,  as  we  see 
further  on,  that  its  end  sides  always  intersect  upon  the  result- 
ant of  the  forces  ;  its  depth  is  always  proportional  (for  paral- 
lel forces)  to  the  moment  at  any  point ;  its  area  to  the  moment 


APPENDIX.]  NOTE    TO    CHAP.  II.  3G3 

of  inertia  of  the  forces ;  while,  finally,  in  a  loaded  beam  the  de- 
flection curve  itself  is  but  a  polygon  or  frame  of  this  character, 
when  the  curve  of  loading  follows  the  law  of  the  moments  in 
the  beam. 

It  is  upon  this  polygon  and  its  properties  that  the  entire 
system  of  Graphical  Statics  is  based. 


364  NOTE   TO   CHAP.  V.  [APPENDIX. 


NOTE  TO  CHAPTEE  V.,  AET.  51. 

15.  In  Fig.  YIIL  (Appendix)  we  have  given  the  construction 
referred  to  in  Art.  51  of  the  text  for  a  system  of  loads  of  given 
intensities.  The  span  s0  s0  is  supposed  to  shift  to  sx  s1?  s%  s2,  etc., 
and  a  certain  cross-section  &0  to  shift  with  it  to  Tc^  &%,  etc.  The 
intersections  of  the  respective  closing  lines  with  verticals 
through  &0,  &u  &2>  etc-?  gives  us  a  curve  between  which  and  the 
polygon  the  greatest  ordinate  gives  the  maximum  moment  for 
the  assumed  cross-section.  The  place  of  this  ordinate  is  the 
position  of  the  cross-section  from  which  we  determine  the  ends 
of  the  span,  and  thus  have  its  position  with  reference  to  the 
loading  when  the  moment  in  k  is  the  greatest  possible. 

Thus  if  this  greatest  ordinate  is- at  the  angle  YIIL  in  the  Fig., 
the  weight  P8  must  rest  upon  the  cross-section.  The  distance 
then  from  P8  to  the  left  end  of  span  s,  is  the  distance  from  SQ 
to  &o  on  left,  and  to  right  end  of  span  s,  is  the  distance  from 
#o  to  SQ  on  right. 

The  ends  s  and  s  being  thus  found,  perpendiculars  through 
them  determine  the  closing  line  L,  and  the  parallel  to  this  in 
the  force  polygon  gives  the  end  reactions  L  0  and  20  L  for  the 
position  of  span  which  makes  moment  at  Jc  a  maximum. 


APPENDIX.]  PIVOT   SPAN.  365 


NOTE  TO  CHAPTEE  XII.,  ART.  124. 

16«  In  Arts.  120  and  121  we  have  given  the  formulae  and 
principles  necessary  for  the  complete  solution  of  the  pivot  span. 
We  propose  here  to  illustrate  more  fully  their  application  by  a 
simple  example. 

Fig.  IX.  represents  such  a  structure.  The  two  outer  spans 
A  B  —  C  D  =  40  f t.  The  central  or  turn-table  span,  B  C  = 
20  ft.  Centre  height  at  B  and  C  =  10  ft.  End  height  =  6  ft. 
Panel  length,  10  ft. ;  each  apex  live  load,  10  tons,  or  1  ton  per 
foot.  Dead  load,  half  'as  much.  Two  systems  of  triangulation, 
as  shown  in  the  Fiff. 

O 

Our  proportions  are  taken  for  the  sake  of  illustration  merely, 
and  not  as  an  example  of  actual  practice.  All  the  points  to  be 
observed  are,  however,  illustrated  as  well  as  by  a  much  longer 
span,  and  more  usual  proportions. 

It  is  to  be  observed  that  the  end  verticals  are  compression 
members  only,  and  cannot  take  tension.  This  is  necessary  -to 
prevent  ambiguity  as  to  the  way  in  which  the  strains  go.  A 
negative  reaction  might  otherwise  cause  tension  in  1  2,  and 
compression  in  F,  or  tension  in  1  5,  compression  in  5  6,  and 
tension  in  A.  If  1  5  cannot  take  tension,  we  have  but  one 
course  for  the  strains,  and  the  problem  is  determinate. 

We  also,  for  similar  reasons,  construct  the  centre  span  so 
that  the  diagonals  take  tension  only,  and  the  verticals  compres- 
sion only.  These  points  as  to  construction  being  settled,  let  us 
proceed,  first,  to  determine  the  reactions. 

1st.  REACTIONS. 

We  shall  consider  the  case  of  the  "  Tipper"  or  secondary 
central  span  only  [Art.  120],  as  this  case  most  nearly  ap- 
proaches the  true  state  of  things.  The  method  of  procedure 
for  fovrjixed  supports  is  precisely  similar,  only  taking  the  for- 
mulse  for  that  case  from  Art.  122. 

The  less  the  span  B  C,  the  nearer  the  case  approaches  to  three 
fixed  supports ;  and  when  the  distance  B  C  is  zero,  n  is  zero,  and 
our  formulas  are  the  same  as  for  beam  over  three  supports. 

For  a  load  in  the  left  span  distant  a  from  A,  these  formulae 
are  as  follows  [Art.  120]  : 


366  NOTE    TO    ART.  124.  [APPENDIX. 

RA  =         j  2H  -  (10  +  15  w,  +  3  n>)  Jc  +  (2  +  n)ji\, 

C  ^g  |(6  +  9 
" 


RD=2H 


1 
J 

I  (2  +  ri)k*  —  (2  -f  3  n  +  3  ft2)  k  I, 


in  which  Jc=^  I  =A  B=C  D,  n  I  =  B  C  and  H  =  4  -f  8  7*  +  3  ft2. 
£ 

We  have  first  to  put  these  formulse  into  the  most  convenient 
shape  for  use  in  the  particular  case  under  consideration.     Thus 

"1  Q  K 

in  this  case  I  =  40,  n  I  =  20 ;  hence  n  =  -  and  H  =  -— ,   and 

2  .          4 

Jc  =  — ,  where  a  has  the  successive  values  of  10, 20, 30, 40  for  P15 

Pa,  P3,  P4.     J&  is  therefore  successively  -,  -,  -  and  -. 
Our  equations  for  reactions  are  then,  after  reducing, 


J  45  £-10  ¥  I 

[1 
1A7i8  17^"  I 
107,  17^J 


Now,  as  we  may  notice,  the  denominator  of  &  is  always  4,  of 
¥  always  64 ;  the  numerator  only  changing  according  to  the 
position  of  the  weight.  These  equations  can  then  be  written 


RA  -  224 


^J2240-584a+5a3  I 

A  |~72  «-</], 
L  -I 


RD  = 

where  a  has  the  values  1,  2,  3  for  P1?  P2,  P3,  etc. 

These,  then,  are  the  practical  formulas  for  this  case,  and  from 


APPENDIX.]  PIVOT   SPAN.  367 

them  we  can  easily  find  the  reactions  for  the  apex  loads  of  10 
tons  each. 

Thus,  for  P!  make  a  =  1,  and  we  have 

RA  =  7.415,      RB  =  RC  =  1.58,      RD  =  —  0.584 

For  Pa  make  a  —  2,  and 

RA  =  4.964,      RB  =  Rc  =  3.035,      RD  =  —  1.035. 

For  P3  make  a  =  3,  and 

RA  =  2.78,        RB  =  Rc  -  4.22,        RD  =  -  1.22. 

For  P4  make  a  =  4,  and 

RA  =  lj  -^B  —  RC  ==  5,  RD  —  —  !• 

Loads  upon  the  centre  of  the  span  B  C  acting,  that  is,  at 
apex  10,  give  no  reactions,  but  are  supported  directly  by  the 
turn-table.  Hence,  for  P5 ;  RA,  RB,  Rc  and  RD  are  zero.  For 
the  first  load,  P7  to  the  right  of  C,  the  reactions  at  A  and  B  are 
the  same  as  for  P3  at  D  and  C,  already  found.  For  the  next 
load,  P8,  the  reactions  at  A  and  B  are  the  same  as  for  P2  at  D 
and  C,  already  found.  For  P9,  the  same  as  for  Pt.  For  P6,  as 
for  P4,  etc. 

We  thus  have  the  reactions  at  A  and  B  due  to  every  indi- 
vidual apex  load,  and  can  now  proceed  to  find  the  strains. 

Our  formulae,  it  will  be  observed,  thus  become  very  simple 
and  easy  of  application  for  any  particular  case. 

2c7.  FLANGES — BRIDGE  SHUT. 

Let  us  first  find  the  strains  in  the  flanges.  We  have  only  to 
apply  the  method  of  moments,  and  the  work  is  so  simple  that 
an  example  or  two  will  suffice. 

We  repeat  again  the  rule.  Conceive  a  section  cutting  only 
three  strained  pieces.  Take  the  intersection  of  two  of  these  as 
the  centre  of  moments  for  finding  the  strain  in  the  third.  The 
moment  of  the  strain  in  this  last  about  this  point  must  be  equal 
to  the  algebraic  sum  of  the  moments  of  all  the  forces  acting 
between  the  section  and  one  end.  Take  Px  for  example.  Its 
upward  reaction  at  A  is  7.415.  [A  negative  reaction  acts 
down.  Thus,  for  P7  above,  the  reaction  at  A  is,  from  our  for- 
mulae, —  1.22.  The  minus  sign  indicates  that  the  reaction  is 
down,  and  that,  neglecting  the  dead  load,  the  girder  must  be 
held  down  to  the  support  A.  If  the  reader  will  draw  roughly 
the  curve  of  deflection,  he  will  see  that  this  is  so.] 


368  NOTE    TO    AKT.  124:.  [APPENDIX. 

Conceive  a  section  through  the  girder  at,  say,  the  centre  of 
flange  A.  It  cuts  4  pieces,  but,  since  the  weight  Pt  acts  only 
through  its  own  system  of  diagonals,  only  three  are  strained. 
The  point  of  moment  for  A  is  then  at  6,  the  intersection  of  the 
other  two  strained  pieces.  The  strain,  then,  in  A  x  by  its  lever 
arm  =  7.415  x  10.  The  lever  arm  of  A  is  6.965  ;  hence 

A  x  6.965  =  7.415  x  10, 
or  A  =  +  10.64  tons  compression, 

because  the  upward  reaction  acting  with  6  as  a  centre  of  rota- 
tion tends  to  compress  A. 

This  strain  evidently  acts  through  both  A  and  B,  since  both 
these  flanges  are  included  by  the  two  diagonals  of  the  system 
for  P!  ;  hence  also,  B  =  +  10.64  tons. 

For  flanges  C  and  D,  since  78  is  the  strained  diagonal,  8  is 
the  centre  of  moments.  The  same  reaction  acts  now  with  the 
lever  arm  30  to  cause  compression,  and  Px  acts  with  the  lever 
arm  20  to  cause  tension.  We  have  then 

C  x  8.955  =  +  7.415  x  30  -  10  x  20, 
or  .        C  =  D  =  2.5  tons  compression. 

Now  we  come  to  the  centre  span,  and  must  carefully  observe 
the  following  points.  Since  D  has  been  found  to  be  compres- 
sion for  Pt,  we  see  at  once  that  the  whole  upper  flange  for  the 
span  A  B  is  for  this  weight  in  compression.  Diagonal  8  9  is 
therefore  in  tension.  Were  there  no  vertical  strut  at  B,  this 
would  cause  compression  in  910.  But  brace  910  cannot  by 
construction  take  compression.  The  strained  pieces  cut  by  a 
section  through  E  are  then  E,  B  11  and  K,  which  give  us  the 
centre  of  moments  at  B  for  strain  in  E.  Observe,  that  were 
it  not  for  the  vertical,  we  should  have  had  10  for  the  centre 
of  moments;  or,  with  the  vertical,  had  D  been  found  tension, 
8  9  would  have  been  compression  ;  there  would  then  have  been 
no  strain  in  the  vertical,  that  being  incapable  of  tension,  and 
diagonal  9  10  would  have  been  strained,  thus  giving  us  also  10 
for  the  centre  of  moments.  Attention  to  the  above  is  necessary 
in  order  to  properly  pass  from  the  span  A  B  into  the  middle 
span. 

We  have  then  for  strain  in  E 

E  x  10  =  7.415  x  40  -  10  x  30,  or  E  =  -  0.34, 


APPENDIX.] 


PIVOT    SPAN. 


369 


or  in  tension,  as  indicated  by  the  sign,  since  the  moment  of  P, 
overbalances  that  of  the  reaction. 

\Note. — The  different  lever  arms  are  easily  obtained  from 
the  known  dimensions  of  the  truss.  We  have  considered  it 
unnecessary  to  detail  how  they  are  to  be  found.  They  may 
either  be  measured  to  scale  from  the  frame  or  computed  trigo- 
nometrically.] 

The  lower  flanges  are  found  in  similar  manner. 

Thus,  strain  in  F  is  zero,  since  it  passes  through  the  point  of 
moments. 

For  G  and  H,  we  have 

G  x  8  =  -  7.415  x  20  +  10  x  10,  or  G  —  -  6.04  tension. 
In  like  manner,  for  I, 

I  x  10  =  -  7.415  x  40  +  10  x  30  =  +  0.34. 

For  K,  for  similar  reasons  as  above  for  E,  we  have  centre  at 
11,  and  therefore  the  reaction  at  B  also  enters  into  the  equa- 
tion of  moments,  and 

K  x  10  =  -  7.415  x  50  +  10  x  40  —  1.58  x  10,  or  K  =  +  1.34. 
We  have  then,  finally,  for  the  strains  in  the  flanges  due  to  Pt 


A 

B 

C 

D 

E 

F 

G 

H 

I 

K 

Pi 

+10.64 

+10.64 

+2.5 

+2.5 

-0.34 

0 

-6.04 

-6.04 

+0.34 

+1.34 

In  a  precisely  similar  manner  we  find  the  strains  due  to  P2, 
P3  and  P4. 

We  have  only  to  observe  that  for  P7,  the  first  weight  to  the 
right  of  C  in  the  other  span,  the  reaction  at  A  is  negative  and 
equal  to  the  reaction  of  P3  at  D,  already  found,  or  —  1.22. 

Now  as  we  suppose  the  end  A  bolted  down,  this  reaction  acts 
as  a  weight  of  1.22  tons  suspended  from  the  end.  So  for  the 
reactions  of  P8  and  P9,  viz.,  —  1.035  and  —  0.584.  These  reac- 
tions, moreover,  must  all  take  effect  through  diagonal  1  2  and 
flange  F,  as  the  end  vertical  cannot  take  tension. 

Finding  then  the  strains  due  to  each  of  the  other  weights, 
we  can,  finally,  tabulate  our  results  as  on  next  page : 
24 


370 


NOTE   TO    ART.  124. 


[APPENDIX. 


STRAINS    IN   FLANGES LIVE   LOAD BRIDGE   SHUT. 


A 

B 

0 

D 

E 

F 

G 

H 

I 

K 

PI 

+10.64 

+  10.64 

+  2.5 

+  2.5 

—  0.34 

0 

—  ?.r 

—  6.04 

—  6.04 

+  0.34 

-1.34 

pa 

0 

+  12.47 

+  12.47 

—  0.14 

—  0.14 

—  7.1   j—  5.32 

—  5.82    +2.14! 

P3 

+  3.9 

+  3.9 
+  2.5~ 

+  9.31 

+  9.31 

+  1.12 

0 

—  6.95     —  6.95 

—  1.12 

+1.88 

P4 
P6 

0 

+  2.5 

+  4.0 

+  4.0 

—1.42 

—  1.42     —  3.;£i 

—  3.33 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

P6 

0 

—  2.5 

—  2.5 

—  4.0 

—  4.88 

—  4.0 

+  1.42 

+  1.42     +  3.33 

+  3.33 

0 

PT 

0 
~~0 

—  3.1  |  —3.1 

—  4.88 

+  1.74 

+  1.74 

+  4.U6 

+  4.U6 

+l.bb 
+2.14 

P8 

-2.6 

-2.6 

—  4.4 

—  4.14 

+  1.46 

+  1.46 

+  3.45 

+  3.45 

P» 

0 

—  1.4 

—  1.4 

—  2.34 

—  2.34 

+  o.as 

+  0.83 

+  1.95 

+  1.95 

+1.34 

Total 
Strains 

+14.54 

+  29.51 

+  26.78 

+  15.81 

+  5.12 

+  5.45 

+  5.45 

+  12.79 

+  13.13 

+9.38 

; 

-9.6 

-  9.6 

—  15.76 

—  15.84 

-  8.52 

—  21.61  1  —  21.64 

—  9.77 

-1.34 

In  the  two  horizontal  lines  at  bottom,  we  have  the  total 
strains  of  each  kind  caused  by  the  live  load. 

3d.  FLANGES — BRIDGE  OPEN — DEAD  LOAD. 

We  have  next  to  find  the  strains  due  to  the  dead  load  when 
the  span  is  open. 

We  have  then  5  tons  at  each  apex,  except  the  ends,  where 
we  have  P0  —  2.5  tons. 

These  strains  are  easily  found  by  moments  as  above,  and  we 
have  then  the  following  table : 


A 

B 

C 

D 

E 

F 

G 

H 

I 

K 

PO 

0 
0 

—  6.2 

—  6.2 

—  10.1 

—  10.1 

+  3.57 

+  3.57 

+  8.3 

+  8.3 

+  10 

P> 

0 

—  11.1 

—  11.1 

—  15.0 

0 

+  6.1 

+  6.1 

+  15.0 

+  15.0 

P2 

0 

0 

0 

—  10.1 

—  10.0 

0 

0 

+  5.5 

+  5.5 

+  10.0 

P3 

0 

0 

0 

0 

—  5.0 

0 

0 

0 

+  5.0 

+  5.0 

Total 
Strains 

0 

—  6.2 

—  17.4 

—  31.2 

—  40.0 

+  3.6 

+  9.7 

+  20.1 

+  33.8. 

+  40.0 

If  now,  as  should  be  the  case,  we  suppose  the  centre  sup- 
ports raised  above  the  level  of  the  ends,  so  that  the  ends  just 
bear,  then  these  strains  above  act  even  when  the  bridge  is  shut. 
As  we  have  already  seen  in  Art.  131,  our  formulae  for  the 
reactions  are  not  affected  by  this  state  of  things.  The  strains 
due  to  live  load  will  then  be  increased  by  those  above,  and  we 
thus  have  for  the  total  maximum  strains  which  can  ever  occur, 


APPENDIX.] 


PIVOT   SPAN. 


371 


A 

B 

0 

D 

*    E 

F 

a 

H 

I 

K 

+  14.5-1 

+  29.51 

+  26.78 

+  15.81 

+  5.12 

+  9.05 

+  15.15 

+  32.89 

+  46.93 

+  49.  S8 

—  15.8 

—  27.0 

—  46.96 

—  55.84 

—  8.52 

—  21.51 

—  21.64 

-9.77 

—  1.34 

Of  course,  for  this  condition  of  things  the  ends  must  always 
be  bolted  down. 

It  is  sometimes  customary  to  raise  the  ends  by  an  apparatus 
for  that  purpose,  after  closing  the  draw,  until  the  proper  pro- 
portion of  the  dead  load  takes  effect  also  as  a  positive  reaction. 

We  can  easily  find  the  strains  in  this  case  also  by  adding 
the  numbers  in  the  last  horizontal  line  of  our  table  for  bridge 
shut,  with  their  proper  signs,  and  taking  half  the  results  for  a 
new  line  for  dead  load  strains.  The  resulting  strains  can 
then  be  found  precisely  as  in  the  table  of  Art.  8  (Appendix). 
We  must  also  find  the  strains  for  bridge  open  as  above,  and  then 
take  the  greatest  strains  of  each  kind  from  these  two  tables. 

In  this  case  'the  strains  would  be  differently  distributed. 
Flange  E  will  be  always  in  tension,  A  and  K  always  in  com- 
pression ;  the  compression  in  B  C  and  D  will  be  somewhat 
greater  than  above,  and  the  tension  in  the  same  flanges  less. 
The  reader  can  easily  deduce  the  strains  for  this  case  from  the 
two  preceding  tables. 

If  the  truss  may  act  as  a  girder  over  four  fixed  supports, 
we  should,  in  order  to  be  certain  of  the  maximum  strains, 
make  the  calculation  for  this  case  also,  using  the  formulae  of 
Art.  122.  This  is  unnecessary,  however,  if  the  supports  B  and 
C  can  never  sink  far  enough  to  strike  the  turn-table,  or  be  im- 
peded in  their  motion. 

4cth.  STRAINS  IN  THE  DIAGONALS. 

We  may  find  the  strains  in  the  diagonals  also  for  each 
weight  separately,  both  for  bridge  open  and  shut ;  and  a  pre- 
cisely similar  method  of  tabulation  will  give  the  strains. 

It  will  here  be  found  preferable  to  make  a  series  of  dia- 
grams, as  illustrated  in  Fig.  86,  Art.  124,  for  each  weight  and 
its  own  system  of  triangulation.  We  obtain  thus  the  diagonal 
strains,  and  at  the  same  time  check  the  results  obtained  for  the 
flanges  above. 

If  we  wish  to  calculate  the  diagonals,  it  will  be  better  to  find 
the  resultant  shear  acting  upon  the  diagonal,  and  multiply  it  by 
the  secant  of  the  angle  the  diagonal  makes  with  the  vertical. 


372  NOTE    TO    ART.  124.  [APPENDIX. 

We  can  also,  if  we  wish,  apply,  the  method  of  moments. 
Thus,  if  we  determine  the  point  of  intersection  in  the  present 
case  of  the  inclined  upper  flange  with  the  horizontal  lower 
flange,  this  point  will  be  a  common  centre  of  moments  for  the 
diagonals.  The  lever  arms  of  the  diagonals  with  reference  to 
this  point  must  next  be  determined,  and  then  we  are  ready. 

This  point  above  for  centre  of  moments  is  easily  found  ;  thus 
4  :  40  : ;  10  :  100. 

It  is  therefore  60  ft.  to  the  left  of  A,  or  100  ft.  left  of  B. 
Take  now  any  diagonal,  as  3  4.  Its  angle  with  the  horizontal 
is  very  nearly  42°,  and  with  the  vertical  48°.  Its  lever  arm  is 
then  80  sin  42°  =  53.5,  and  sec.  of  angle  with  .vertical  is  1,49. 

Now  take  the  weight  P3.  Its  upward  reaction  at  A  is  4.964, 
P2  being  10. 

We  have  then 

[str.  in  3  4]  x  53.5  =  10  x  80  —  4.964  x  60  =  +  502.16. 

The  resultant  rotation  is  then  positive,  or  from  left  to  right. 
The  point  P2  then  sinks  and  4  rises,  and  34  is  in  tension  and 
502.16 

—  "KO~K~  =  ~~  9-38  tons. 
oo.o 

This  is  sufficient  to  illustrate  the  method. 

For  the  first  method  referred  to  above,  viz.,  that  by  resultant 
shear,  the  following  points  are  to  be  observed : 

When  a  piece  slopes  towards  the  nearest  support,  we  say  it  is 
sloped  as  a  strut,  whatever  the  real  strain  in  it  may  be. 

When  it  slopes  away  from  the  nearest  support,  it  is  sloped  as 
a  tie. 

The  simple  shear  is  the  reaction  at  the  support  minus  the 
weights  between  any  point  and  that  support. 

If  any  three  strained  pieces  are  cut  by  a  section  through  the 
structure,  the  strains  in  these  pieces  are  in  equilibrium  with  the 
simple  shear  at  this  section.  Hence  the  algebraic  sum  of  the 
vertical  components  of  these  pieces  must  be  equal  and  opposite 
to  the  shear  itself. 

In  order  to  add  these  vertical  components  with  proper  signs, 
we  must  remember  that  if  a  flange  is  in  tension  and  sloped  as  a 
strut,  or  in  compression  and  sloped  as  a  tie,  we  add  the  vertical 
component  of  the  strain  in  it  to  the  simple  shear  already 
obtained.  If  in  compression  and  sloped  as  a  strut,  or  tension 
and  sloped  as  a  tie,  we  subtract. 


APPENDIX.]  PIVOT    SPAN.  373 

The  resultant  shear  thus  obtained  then,  multiplied  by  the 
secant  of  the  angle  with  vertical,  gives  the  strain  in  diagonal. 

If  the  sign  of  the  result  is  negative  (— ),  it  shows  that  the 
strain  on  the  diagonal  is  contrary  to  that  indicated  by  its  slope. 

To  illustrate,  let  us  again  take  the  weight  P2  and  consider 
diagonal  34. 

The  simple  shear  at  apex  4  is  4.964  —  10  =  —  5.036.  The 
strain  in  C  for  Pa  we  have  found  to  be  compression,  and  equal 
to  -h  12.47.  It  is  sloped  as  a  strut,  and  its  vertical  component  ia 
therefore  to  be  subtracted  from  the  shear  above.  Since  its 
angle  is  nearly  5°  43'  with  the  horizontal,  this  vertical  com- 
ponent is 

12.47  x  sin  5°  43'  =  1.24. 

Since  H  is  in  this  case  horizontal,  it  has  no  vertical  com- 
ponent. 

The  resultant  shear  is  then 

-  5.036  -  1.24  :=  -  6.276. 

As  the  secant  of  the  angle  of  3  4  with  the  vertical  is  1.49, 
we  have  for  the  strain  in  34,  •-  6.276  x  1.49  =  —  9.35. 

This  result  being  minus,  and  3  4  being  sloped  as  a  strut, 
the  strain  is  9.35  tons  tension,  agreeing  closely  with  the  value 
found  above  by  moments. 

The  above  method  is  preferable  to  the  method  by  moments 
for  the  diagonals,  as  we  have  only  to  determine  the  secants  for 
the  verticals  and  the  sines  for  the  flanges,  which  is  in  most 
cases  easier  than  to  find  the  lever  arms  for  the  diagonals  and 
the  points  of  intersection  of  the  upper  and  lower  flanges  in  each 
panel.  It  is,  like  the  method  of  moments,  of  general  applica- 
tion to  any  framed  structure  whose  outer  forces  are  known. 

The  method  of  diagram  in  Art.  124  will  be  found  preferable 
to  both. 

It  is  unnecessary  to  pursue  our  example  further.  With  the 
mutual  checks  of  the  two  methods  of  calculation  explained 
above,  as  well  as  the  diagrams,  correct  results  cannot  fail  to  be 
obtained.  The  diagrams  should  always  be  made  first,  as  they 
settle  by  mere  inspection  many  points  which  may  at  first  cause 
trouble — such  as  whether  the  shear  in  a  piece  is  subtract! ve  or 
not  according  to  our  rule,  the  character  of  the  strains  in  dif- 

O  ' 

ferent  pieces,  etc.     It  is  well  to  indicate  on  the  diagrams  com- 
pressive  strains  by  double  or  heavy  lines. 


374 


NOTE   TO   ART.    128. 


[APPENDIX. 


NOTE  TO  AKT.  128,  CHAPTER  XII. 

17.  We  wish  here  to  call  more  particular  attention  to  the 
relative  economy  of  the  continuous  as  compared  with  the 
simple  girder.  This,  we  think,  is  greater  than  is  generally  sup- 
posed. It  may  reach  from  18  to  25,  and  even  as  high  as  50 
per  cent. 

Take  the  example  worked  out  in  Art.  128,  Fig.  88.  We 
have  obtained  the  maximum  strains  in  that  Art.  upon  every 
piece. 

We  give  them  below,  compared  with  the  strains  in  the  same 
pieces  for  a  simple  girder  of  same  dimensions  anb  load : 

Aa         Ac          Ae         Ag        A*         B6         Ed          Bf         BA 
Continuous..— 203.5     +63.6     +115.3     +63.6    —203.5     +89.3    —115.9    —115.9     +89.3 

Simple 0          +180      +240        +180          0          —90       —210       —210       —90 

a&  be          cd          de         ef         fg          gh          hk 

Continuous.. +189.3    —109.9     +109.9     +45.5     +45.5     +109.9     +109.9     +189.3 
Simple +127.3    —127.3     +56.5      +56.5    —56.5     +56.5      —127.3     +127.3 

It  will  be  seen  at  once  that  there  is  a  saving  in  the  flanges — 
about  11  per  cent,  in  all — but  the  bracing  is  heavier,  giving  lit- 
tle or  no  saving.  The  span  is  too  short  to  properly  represent 
the  relative  economy  of  the  two  systems. 

If  we  take  a  truss  such  as  represented  in  PI.  2,  Fig.  VII., 
Appendix,  by  the  full  lines  only,  omitting  the  dotted  verticals  and 
diagonals — height  6  ft.,  span  50  ft.,  panel  length  10  ft,,  dead 
load  5  tons  per  panel,  live  load  7  tons  per  panel — and  calculate 
the  strains  in  the  pieces  for  a  simple  girder,  and  then  as  a  con- 
tinuous girder  of  two  spans  and  three  spans,  we  have  the  fol- 
lowing results : 


1  span 
strain  in  tons. 

2  spans 

3  spans 
end. 

3  spans 
middle. 

205  2 

231.6        . 

229.8 

239.4 

Lower  Chord  . 

200. 

168.7 

152.9 

122.6 

Upper  Chord  .... 

200. 

^158.7 

159.9 

135.2 

Total 

605  2 

559 

542  6 

497  2 

Per  cent,  saving.  . 

8  per  cent. 

10  per  cent. 

18  per  cent. 

APPENDIX.]  THE    CONTINUOUS    GIRDEK.  375 

We  have  then,  in  the  first  case,  a  saving  of  8  per  cent.,  in  the 
second,  10,  and  in  the  third,  18  per  cent,  over  a  simple  girder. 
Quite  a  notable  saving,  although  the  spans  are  very  short,  and 
although,  in  the  first  two  cases  (the  spans  being  end  spans),  we  do 
not  obtain  the  full  advantages  of  continuity.  If,  then,  instead 
of  three  simple  girders  of  the  above  dimensions,  we  should  con- 
struct the  girder  continuous  over  the  piers,  we  should  save  in 
strain,  and  hence  in  material,  10  per  cent,  in  each  end  span,  and 
18  per  cent.,  or  nearly  twice  as  much,  in  the  centre  span. 

The  advantage  of  continuity  is  rendered  still  more  apparent 
by  taking  a  longer  span.  Thus  for  a  girder  of  200  ft.,  height 
20  ft. — 10  panels,  and  double  system  of  triangulation,  similar  to 
Fig.  YII. — for  a  live  load  of  20  tons  per  panel,  and  dead  load 
of  10  tons — we  have  the  following  results : 

One  span.  Two  spans.  Five  spans. 

r  centre. 

Bracing 1398.6  1428.2  1596.2 

Lower  Chord., 2400.  1793.2  1395.7 

Upper  Chord . .  2550.  1981.6  1622.6 

Total 6348.6  5203.0  4614.5 

Per  cent,  saving 18  per  cent.        27  per  cent. 

That  is,  we  have  a  saving  of  18  per  cent,  instead  of  only  8 
per  cent,  as  before,  for  two  spans,  and  of  27  per  cent,  for  the 
centre  span  of  five  spans.  For  three  spans,  then,  of  this  length 
we  should  save  18  per  cent,  on  the  end,  and  at  least  30  per  cent, 
on  the  centre  span. 

If  we  suppose  the  same  girder  as  above  fastened  or  fixed 
horizontally  at  the  ends,  we  shall  have  the  case  of  a  middle  span 
in  a  very  great  number,  and  may  expect  to  find  the  greatest 
saving  possible  for  this  length. 

The  formulae  of  Chapter  XIII.,  as  also  the  simple  graphical 
method  for  this  case,  given  in  Chapter  XII.,  Art.  114  [Fig.  80], 
enable  us  to  solve  this  case  easily.  The  reader  will  find,  on  mak- 
ing the  calculation,  the  following  strains  : 

Strain  in  tons. 

Diagonals 1279.2 

Upper  Chord 940.2 

Lower  Chord 965.4 

Total 3184.8 

Per  cent,  saving 49.8 


376  NOTE    TO    ART.  128.  [APPENDIX. 

The  above  will  serve  to  illustrate  the  point  in  question  quite 
as  well,  perhaps,  as  an  extended  theoretical  discussion.  We  see 
that  the  saving  increases  rapidly  with  the  length  of  the  span, 
and  may  easily  rise  as  high  as  30  or  40  per  cent.,  while  in  some 
cases  even  50  per  cent,  may  be  realized. 


THE   DISADVANTAGES    OF   THE    CONTINUOUS    GIRDER   ABE  I 

1st.  The  fact  that  the  various  pieces,  especially  the  chords, 
undergo  strains  of  opposite  character. 

This,  in  wrought- iron  structures,  we  venture  to  think  of  little 
importance.  The  extra  work  and  cost  of  chords  and  chord  con- 
nections necessary  to  secure  the  flanges  against  both  compressive 
and  tensile  strain,  can  hardly  amount  to  10,  18,  30,  or  even  50 
per  cent,  of  the  cost  of  girder ! 

%d.  Difficulty  of  calculation. 

We  have,  we  trust,  in  what  precedes,  and  in  Chapter  XIII., 
succeeded  in  removing  this  objection. 

The  opinion  is  widespread  among  engineers  that  the  deter- 
mination of  strains  in  the  continuous  girder  is  impracticable 
and  involved  in  mystery.  No  opinion  could  well  be  more  un- 
founded. The  accurate  and  complete  calculation  for  all  pos- 
sible loading,  live  or  dead,  is  precisely  similar  to  and  offers  no 
more  difficulty  than  the  simple  girder  itself. 

The  formulae  for  moments  and  shears  are,  as  we  have  seen, 
simple  and  easy  of  application. 

The  graphic  method  here  developed  offers  also  a  thorough 
solution.  In  view  of  both,  and  of  the  extensive  literature  upon 
the  subject  (which  seems,  by  the  way,  to  have  been  so  generally 
ignored),  we  can  finally  pronounce  the  problem  to  be  fully 
solved. 

3d.  The  changes  of  strain,  unforeseen  and  often  considerable, 
which  a  small  settling  of  the  piers  or  change  of  level  of  the 
supports  may  occasion. 

This,  be  it  observed,  is  only  of  importance  when  the  piers 
settle  after  the  erection  of  the  superstructure.  If  piers  are  to 
be  considered  as  settling  indefinitely,  or  continuously  during  a 
succession  of  seasons,  continuous  girders  are  not  to  be  thought 
of.  If,  however,  as  is  generally  the  fact,  the  piers  take  their 
permanent  set  during  the  first  season,,  and  afterwards  are  im- 
movable, the  above  objection  has  no  weight.  It  is  not  necessary 


APPENDIX.]          SUPPOETS  OUT  OF  LEVEL.  377 

that  the  piers  should  be  exactly  on  level  or  even  on  line,  or  even 
that  the  differences  of  level  be  known. 

As  shown  in  Art.  121,  these  differences  produce  no  effect, 
provided  the  girder  be  built  to  the  profile  of  the  supporting 
points. 

If  in  any  case  these  differences  are  required,  and  it  is  con- 
sidered difficult  to  determine  them  over  water  with  sufficient 
accuracy,  then  the  proper  reactions  at  the  several  piers  may  be 
weighed  off*  and  the  girder  thus  left  in  position  under  pre- 
cisely the  circumstances  for  which  it  has  been  calculated. 

THE   PRINCIPAL   ADVANTAGES    OF   THE   CONTINUOUS    GIRDER   ARE  : 

1st.  Ease  of  erection,  where  false  works  are  difficult  or  ex- 
pensive. The  girder  may  be  built  on  shore,  and  then  pushed 
out  over  the  piers. 

2d.  Saving  in  width  of  piers,  as  compared  with  width  re- 
quired for  separate  successive  spans.  The  girder  may  be  placed 
upon  'knife  edges  at  the  piers.  In  fact,  such  a  construction  is 
preferable,  as  better  ensuring  the  calculated  strains.  Width  of 
piers  is  undesirable. 

3d.  Saving  in  jnaterial — usually  from  25  to  30  per  cent. 

18.  Continuous  Crirder— Supports  not  on  a  level.  —  In 
Chapter  XIII.  we  have  all  the  formulae  required  for  the  solution 
of  the  continuous  girder  for  supports  on  a  level,  or  all  on  line, 
when  the  deviation  from  level  is  small,  whatever  may  be  the 
number  or  relative  length  of  the  spans.  If  for  a  continuous  girder 
of  constant  cross-section,  the  supports  are  properly  lowered  after 
the  girder  is  placed  upon  them,  we  may  obtain  a  saving  of  23 
per  cent.,  or  more  in  material  over  the  same  girder  with  sup- 
ports all  on  level.  If,  however,  the  cross-section  varies  accord- 
ing to  the  strain — in  other  words,  if  the  girder  is  of  constant 
strength — no  advantage  is  gained  from  thus  lowering  interme- 
diate supports.  Such  disposition  of  the  supports  may  even  act 
injuriously. 

The  formulae  for  shear  and  moments  which  we  have  given 
are,  indeed,  based  upon  the  hypothesis  of  constant  cross-section, 
but  the  strains  in  every  piece  of  the  girder  being  found  for  the 
shears  and  moments  thus  obtained,  each  piece  is  proportioned 

*  An  idea  first  suggested  by  Clemens  Herschel,  C.E.:  Continuous,  Revolving 
Draw  Spans.  Little,  Brown  &  Co. ,  Boston,  1875. 


378  NOTE    TO    ART.    128.  [APPENDIX. 

to  its  strain,  and  the  actual  girder  erected  is  not  of  constant 
cross-section,  but  more  nearly  one  of  uniform  strength.  Formu- 
lae for  the  case  of  supports  out  of  level,  as  well  as  determina- 
tions of  the  best  differences  of  level,  are  hence  of  but  little  prac- 
tical importance,  and  have  not  been  given.  If,  however,  it  be 
required  to  find  the  effect  due  to  the  sinking  of  any  one  pier, 
the  following  may  be  found  of  service. 

Let  the  nih  support  be  depressed  below  the  level  of  the  others 
by  the  distance  A.  Then  the  moments  at  all  the  supports  are 
changed.  The  moments  at  n  and  at  each  alternate  support 
from  n  are  diminished,  and  at  the  others  increased. 

Let  H  =  ?=i 


»  Then,  when  all  the  spans  are  equal,  the  following  formulae 
give  the  moment  at  any  support  : 

1st.  All  spans  equal,     n  —  number  of  lowered  support,  from 
left, 

when  m<n,       Mm  = 


when  m=n,       Mm  =  -  5  -  °n~l  C*-»+2  +  °n  G*-«+1  H, 


when  m  >  n,       Mm  =       HLtl      H, 

CB+l 

where   ^  =  0,    cz  =  +  1,    c3  —  —  4:,    c4  =  +  15,    c5  =  —  56, 
c6  =  +  209,  etc. 

From  the  moments  at  the  supports  the  shears  can  be  readily 
determined  from  the  formula  of  Art.  148,  viz.  : 


Mm+l  a,       _Mm-Mm_1 

- 


where  q  =  P  (1—^)  for  concentrated  load  and  —  r;  for  uniform, 


2d.  Spans  att  unequal. 

when  m  <  n,   cm\  ^-°+i  ~  ^-n+a  +  »s_J+8  —  «»-n+2  I 
L  4  ^-'          J 


APPENDIX.]          SUPPORTS  OUT  OF  LEVEL.  379 


when  m  = 


+  *-« 


4  4-1  ^  (4—1  +  4) 

[Cn-1~gn  +  ^+1~Cn1  6AnEI, 
4-i  4 


s—  m+2 

when  m  >  n*  M    =  7  n  /7 

i_i  <v-i  +  2  ( 

in  which  expressions 


—   —       -  ^  -  ,  4  -  —  7    7 

=_2o4^-c^,etc. 

64  64 

_  A         /7—1          /7    -  9  ^s  +  4-1 

=  U,       <^2  =  Ij       ^3  —   —  ^  -  7  --  > 

^a-1 

_  4:  ft  +  Cl)    (Cl  +  ^-^s-l 

—   --  7  -  7—  —  J 

V-l  ^&-2 

=  _  2  <Z4  ^  +  ^  -  d.         etc. 


The  reader  who  has  learned  the  use  of  the  formulae  of  Chap- 
ter XIII.  will  have  no  difficulty  in  applying  the  above  to  any 
particular  case.  In  the  same  way  as  there  explained,  by  mak- 
ing 4  and  4  zero,  we  may  fix  the  girder  at  the  ends,  etc.  The 
formulae  for  shear  at  any  support  are,  of  course,  the  same  as 
before  (Art.  150). 

Ex.  1.  Let  a  beam  df  two  equal  spans  be  uniformly  loaded 
throughout  its  whole  length,  and  let  the  centre  support  be  low- 
ered by  an  amount  h%  =  *  .  What  are  the  moments  and 

reactions  f 

The  moments  due  to  the  full  load  alone  before  the  support  is 

lowered  are  M!  =  0,    M2  =  — ,     M3  —  0    (Art.  150).      For 

8 

the  moment  due  to  the  lowering  of  the  support  alone,  we  have 
from  the  above  formulae,  since 

J  "  J  — —  — — •  y 


Hence  the  total  moment  is 

.__       w  Z2      w  P     wfi 

IyL0  —  _   —  _    —  _ 

8         16  "  "  16  ' 


380  NOTE    TO    ART.    128.  [APPENDIX. 

For  the  shears,  then,  we  have 

M2      wl       7  ,        9 


7  18  7 

Hence,     R1=  ~  w  I,    R2  =  —  w  I,    RB  =  --w  I. 
ID  lo  lo 


Ex.  2.  Zfow  mw?A  m^stf  w<2  lower  the  second  support  in  the 
above  example,  in  order  that  the  reaction  at  the  centre  support 
may  he  just  zero  f 

In  this  case  we  have 


_wP_  H__  <z0P      3 

~~          ~Z:~ 


g/  _  2  M2          ,  _  wl  __  6  A2  E I  5      7      6  AS  E I 

T~  "  T  Z3  ~  4  ^  ~~p     " 

If  this  is  to  be  zero,  we  have 

_  5wl*      and    M  =  _  1  ^  ? 

and  R!  =  w  Z,     Rg  =  0,    R8  =  w  I, 

or  precisely  as  for  a  beam  of  single  span  ^and  length  2  I. 

Ex.  3.  J.  beam  of  four  equal  spans  is  unloaded,  and  the 
third  support  is  lowered  by  an  amount  h%  =       _    . 
the  reactions  f 

Ans.    Ri  = 


Ex.  4.  ^i  J^^m  of  five  equal  spans  rests  as  a  continuous  gir- 
der over  six  supports.  Having  given  the  dimensions  of  the 
beam,  length  of  spans,  and  coefficient  of  elasticity  ;  to  deter- 
mine the  reactions  due  to  a  sinking  of  the  third  support  one- 
eighth  of  an  inch. 

Let  the  beam  be  of  wood,  1  foot  wide,  1.5  deep, 
I  =  20  feet,     5  =  5,      r  =  3,     E  =  288,000,000  Ibs.  per  sq.  ft., 
A,  =  £  in.  =  0.010417  ft. 


APPLNDIX.]          SUPPORTS  OUT  OF  LEVEL.  381 

Then,      c,  =  1,     c3=  -  4,     c,  =  15,     £5  =  -  56,     c6  =  209, 

540EI&,      __  906EI^  _576EU3 

and    M2  =       209  P     '     M3=  209  /*     '    M4-~~~~ 


1  3  375 

or,  inserting  the  constants  above,  and  I  =  —  b  dB  =    '       , 

M!  =  M6  1=  0,  M2  =  5448,  M3  =  -  9142,  M4  =  5812, 
M5  =  -  1453  ft.  Ibs. 

If  all  the  spans  are  unloaded.  For  the  reactions  necessary 
to  bend  it  and  keep  it  down  to  the  supports,  Rt  =  —  272  Ibs., 
112  =  1002,  R3=-1477,  R4  =  llll,  R5  =  -  436,  R,  =  73. 

If  the  beam  weigh  75  Ibs.  per  foot^  what  deflection  of  third 
support  will  raise  the  left  end  from  the  abutment  f 


Ans.  Rt  =       w  I  =  ?  or  7^  =  0.0226  f  t.=0.2712  in. 


It  will  be  observed  that  a  small  difference  of  level  has  then 
considerable  effect. 

Ex.  5.  Two  equal  spans  are  uniformly  loaded.  How  high 
must  the  centre  be  raised  in  order  that  the  ends  may  just 
touch  f 

This  is  the  case  of  the  pivot  span  with  centre  support  raised. 
(See  Art.  121.) 

The  reactions  at  the  ends  are  zero.  At  pier,  then,  Rg  =  2  w  I, 
hence  moment  at  pier  JYT2  =  \  w  P.  But  the  moment  when  the 
supports  are  on  level  is  M2  =  -J-  w~P,  hence  f  wff  must  be  due 
to  the  elevation  of  the  support.  Then  from  our  formulae, 

3  E  I  ^  ,  w  P 

-—  ,    or     A,=    - 


which  is  precisely  the  same  as  the  deflection  of  an  horizontal 
beam,  fastened  at  one  end,  and  free  at  the  other  (Supplement 
to  Chap.  VII.,  Art.  13).* 

*  The  calculations  and  formulae  given  in  this  Note  (pp.  374-381)  are  taken, 
by  permission,  from  a  manuscript  work  on  the  Theory  and  Calculation  of  Con- 
tinuous Bridges,  by  Mansfield  Merriman,  C.B.,  which,  we  hope,  will  be  soon 
given  to  the  public. 


382  NOTE   TO    CHAP.  XIV.  [APPENDIX. 


NOTE  TO  CHAPTEE  XIV. 

THE   BRACED    AKCH. 

19.  The  subject  of  braced  arches  is  an  important  one,  and 
is  treated  in  no  work  with  the  fulness  and  completeness  it 
deserves.  The  methods  and  formulae  of  Chapter  XIY.  will, 
we  believe,  render  the  determination  of  the  strains  in  this  class 
of  structure  easy,  and  we  propose  in  the  following  to  illustrate 
their  use,  so  far  as  may  be  necessary  to  render  their  application 
clear. 

In  PI.  4,  Fig.  X.,  we  have  represented  a  braced  circular 
arch  with  parallel  flanges.  Span  of  centre  line  =  175  ft. ; 
radius,  201.4  ft. ;  rise,  20  ft.  In  practice,  the  panels  would  be 
taken  of  equal  length ;  for  convenience  of  calculation,  however, 
we  suppose  the  panel  length  to  vary  so  that  the  horizontal  pro- 
jection is  constant,  and  equal  to  25  ft.  Depth  of  arch,  10  ft. 
Hence,  span  of  lower  flange  =  170.6  ft. ;  rise,  19.5  ft. ;  radius, 
196.4  ft.  Span  of  upper  flange,  179.34  ft. ;  rise,  20.5  ft. ;  radius, 
206.4  ft. 

Since  the  flanges  are,  in  practice,  broken  lines,  and  not  true 
curves,  the  depth  or  lever  arm  for  upper  flanges  is  9.43  ft.,  for 
lower  flanges,  10.4  ft. 

The  determination  of  the  other  dimensions  required  is  then 
easy,  and  a  simple  question  of  trigonometry. 

Thus  we  have  for  the  half  central  angle  a  =  25°  45',  and  for 
the  distances  of  the  apices  from  the  chord  of  the  centre  line : 

For  1...— 4.5        3....   4.7        5. ...11.3        7 14.6ft. 

"    2.... 10.8        4.... 18.5        6.... 23.4        8.... 25      " 
We  suppose  the  load  at  each  apex  10  tons,  and  shall  consider 

~Lst.  Arch  hinged  at  crown  or  ap6x  8,  and  at  the  ends  of 
the  lower  flange — the  flanges  H  and  A  being  removed. 

2d.  Arch  hinged  at  apex  8,  and  at  the  ends  of  the  centre 
line — the  flanges  A  and  E  butting  against  a  skew  ~back  or  pivoted 
plate,  and  the  flange  H  only  being  removed. 

3d.  Arch  continuous  at  crown — the  flange  H  being  retained, 
and  hinged  at  ends  of  lower  flanges. 


APPENDIX.]  THE  BRACED  ARCH.  383 

kth.  Arch,  as  in  3d,  but  pivoted  at  ends  of  centre  line. 

5th.  Arch  without  hinges,  or  continuous  at  crown  and  fixed 
at  abutments. 

These  cases  will  illustrate  all  the  principles  of  Chapter  XIY., 
and  a  comparison  of  the  results  obtained  in  each  case  may 
prove  instructive. 

2O.  Arch  hinged  at  apex  8,  and  at  the  extremities  of  the 
lower  flange—  flanges  H  and  A  being  removed. 

From  Art.  158  we  can  easily  find  the  reaction  and  horizon- 
tal thrust  at  left  end  either  by  construction  or  formula  for 
every  weight.  Thus 


For  the  first  weight  P,,  then, 


For  the  weight  P1 


In  similar  manner,  we  find 

P!  =  0.603,  P2  =  1.33,  P3  =  2.06, 

H!  =  2.74,  H2  =  3.84,  H3  =  5.9, 

P4  =  2.8,  P5  =  3.53,  P6  =  4.2, 

H4  =  8.1,  H5  =  10.2,  H6  =  12.1, 

P7  =  5.0,  P8  =  5.7,  P9  =  6.47,  P10=7.2, 

H7  =  14.4,  He  =  12.1,  H9  =  10.2,  H10=  8.1, 

Pu=7.94r,          Pu=S.6,  PIS-  9.4, 

Hu=  5.9,  H12=  3.84,         H13=  1.74. 

It  will  be  at  once  seen  that  the  reaction  of  P8  at  A  is  the 
same  as  of  P6  at  B,  or  equal  to  10  —  P6  ;  while  the  horizontal 
thrust  is  the  same  for  both.  We  need  them  only  to  find  P  and 
H  for  weights  1  to  7,  and  can  then  at  once  write  down  the 
others.  We  are  now  ready  either  to  calculate  or  diagram  the 
strains. 


384:  NOTE    TO    CHAP.  XIV.  [APPENDIX. 

Thus,  for  instance,  for  P10  (see  Fig.  X.),  we  lay  off  the  reaction 
at  A  upwards  to  scale  from  C  to  D,  then  the  horizontal  thrust 
at  A  from  D  to  1,  then  the  equal  thrust  at  B  from  1  back  to  D, 
then  the  reaction  at  B  from  D  to  8,  and  finally  the  weight 
down  from  8  back  to  C,  thus  closing  the  polygon  for  the  exterior 
forces.  Lines  parallel  to  the  pieces  then  give  the  strains. 
Thus  the  thrust  and  reaction  at  A  are  in  equilibrium  with  E 
and  1  2.  Then  1  2  is  in  equilibrium  with  B  and  2  3,  and  so  on. 
Observe  that  the  diagram  checks  itself.  Thus  the  last  diagonal 

7  8  must  be  in  equilibrium  with  6  7  and  G  (flange  H  being  re- 
moved), and  that  this  is  so  is  shown  by  the  strain  in  7  8  passing 
exactly  through  8,  thus  making  the  strain  in  H  zero.     We  can 
also  check  the  work  by  calculating  the  strain  in  the  last  flange 
D  by  moments.     Thus  for  P10 

D  x  9.43  =  7.2  x  72.8  -  8.1  x  19.1  -  10  x  25  =  119.45, 
or  D  =  +  12.6. 

If  this  agrees  with  D  as  found  by  diagram,  and  if  the  diagram 
also  checks,  we  may  have  confidence  in  the  accuracy  of  the 
work,  and  at  once  scale  off  the  strains.  Observe  that  diagonals 
45  and  56  are  both  tension ;  also  that  F  and  G  are  tension. 

We  have  given  also  the  diagram  for  Pn,  which  the  reader 
can  easily  follow  through  for  himself.  F  and  G  are  both  ten- 
sion, 3  4  and  4  5  both  compression.  The  horizontal  thrust  is 

8  &,  and  the  reaction  at  A  =  b  1. 

We  thus  make  a  diagram  for  each  separate  weight,  and  then 
taking  the  dead  load  at  -J  the  live,  we  can  form  the  following 
table  of  strains.  Since  we  wish  only  the  maximum  strains  on 
one-half  the  arch,  those  on  the  other  half  being  precisely  simi- 
lar, we  can  diagram  the  strains  due  to  all  the  weights  upon  the 
right  half  at  once  by  taking  the  sum  of  their  reactions  and 
thrust  at  A.  We  have  then  the  following  table : 


APPENDIX.] 


THE  BRACED  ARCH. 


385 


s 

. 

. 

. 

10 

"* 

. 

o 

o 

"<* 

05 

co 

R 

£ 

• 

• 

; 

7 

7 

1 

Th 

1 

1 

1 

<M 

1 

3 

1 

n 

\ 

Maximuir 

CO 

£ 

+ 

*> 

1 

+ 

C5 

0 

TH 

+ 

J> 

§ 

+ 

0 

£ 

f 

•rH 

oof 

•* 

+ 

10 

+ 

00 

TH 

1O 

+ 

t- 

9 

+ 

00 

TH 

+ 

05 

X 

+ 

Jl-* 

§* 

i> 

05 

O5 

+ 

OS 

5 

+ 

<* 

0 
10 

+ 

CO 

cs' 

CO 

+ 

«*. 

3 

c^ 

! 

CO 

5* 

+ 

«q 

•!—  1 

1 

CO 
idj 

TH 

+ 

00 

as 

1 

TH 

o 

TH 

+ 

J> 

10' 

1 

TH 

05 

1 

T- 

i> 

TH 

o 

o 

•^ 

o 

GO 

10 

10 

1O 

11 

M  g 

00 

• 

o 

TH 

1 

* 

: 

10 

1 

g 

1 

OS 

I 

^ 

1 

05 

1 

$ 

£ 

\ 

(5 

00 

00 

•<* 

^* 

os 

03 

« 

10 

TH 

0 

o 

CO 

11 

g 

+ 

§g 

+ 

8 

+ 

+ 

s 
-+ 

CO 

+ 

$ 

+ 

o 

+ 

?>' 

CO 

+ 

J> 

+ 

05 

CO 

£- 

-h 

s 

+ 

1—  i 

00 

CO 

•* 

c? 

T^ 

0 

TH 

0 

rH 

O5 

CS 

,_, 

M 

p7 

00 

+ 

^h 

05 

+ 

J> 

1 

Tt? 

1 

~ 

rt< 

TH 

+ 

TH 

+ 

CO 

1 

; 

" 

; 

1  . 

Pk 

*>' 

1—  1 

10 

0 

T—  1 

+ 

T-l 

co' 

+ 

CO 
CO 

1 

t^* 

o 

1 

0 
CO* 

1 

0 
05 

+ 

CO 

o 

TH 

1 

0 

1 

s 

+ 

0 

^- 

1 

oq' 

+ 

0 
05' 

1 

I> 

00 

»o 

CO 

w 

10 

TH 

CO 

to 

1C 

CO 

10 

10 

to 

00 

T-t 

+ 

CD 

+ 

OS 

+ 

o 

+ 

^ 

TH 

0 

1 

OS 

+ 

00 

1 

as 

+ 

CO 

+ 

CD 

I 

CO 

+ 

^ 
1 

0 

PU 

z> 

00 

+ 

IO 
O5' 

05 

+ 

o 
05 

1—  1 

+ 

10 

^ 
+ 

0 

z> 

1 

GO 

l> 

1 

J> 

«d 

+ 

o^ 

\ 

GO 

+ 

t 

: 

T£ 

+ 

10' 

1 

0 

o 

1O 

1O 

10 

CO 

TH 

«e 

•rH 

»o 

t- 

TH 

TH 

Ol 

Pi 

so 

+ 

1C 

•t—  1 

+ 

10 

I—I 

+ 

00 

+ 

0 

+ 

o 

TH 

J 

rJH 

+ 

CO 

1 

«0 

+ 

«0 

1 

CO 

+ 

CO 

+ 

t- 

1 

00 

fk 

T—  1 

05 

+ 

o 

OS 

+ 

JO 

as' 

+ 

OJ 

w 

+ 

<N 
J>« 

+ 

CO 

i-T 

1 

^ 

T-H 

+ 

CO 

1 

to 

^ 
+ 

TH 
-4 

I 

Y—  1 

CO* 

+ 

£- 

I> 

\ 

»0 

as 

1 

J> 

to 

CO 

o 

w 

i> 

CO 

TH 

CO 

rH 

CO 

as 

TH 

1 
PH" 

•<^ 

1 

C5 

+ 

i 

+ 

«o 

CO 

+ 

"* 

CO 

+ 

TH 

<M 

+ 

CO 

1 

CO 

+ 

1O 

+ 

TH 

1 

T-« 

fi 

+ 

TH 
TH 

1 

10' 

+ 

CO 
94 

ft 

^ 

CO 

1 

2 

4- 

CO 

id 

+ 

CO 

§;' 

+ 

10 

& 

+ 

o* 

0 

TH 

+ 

CO 

oi 

1 

o 

05 

+ 

CO 
CO 

+ 

TH 

co' 

1 

1O 
00 

+ 

as 

CO' 

1 

05 

TH 
TH 

+ 

w 

0 

Q 

H 

h 

0 

05 

TH 

CO 
<M 

"* 

CO 

10 

"tf 

CO 
10 

t- 

CO 

GO 

t- 

NOTE    TO   CHAP.  XIV.  [APPENDIX. 

21 .  Arch  hinged  at  Apex  §  and  at  the  Extremities  of  the 
centre  Une ;  FIange§  A  being  retained,  and  only  H  re- 
moved.— The  method  of  solution  is  precisely  the  same  as 
before,  the  only  difference  being  that  the  span  is  now  175  ft. 
instead  of  170.6,  and  the  rise  25  ft.  instead  of  29.5.  The  reac- 
tions and  thrust  will  then  be  somewhat  different.  Thus,  for 
the  left  abutment  A, 

P1=0.71,     Pa  =1.42,    P8  =2.14,     P4=2.85,     P5  =    3.57, 
H!=2.48,     H2  =  4.97,  H8  =  7.5,       H4  =  9.97,    H5  =  12.5, 

P6=  4.28,   P7=  5.00,   P8=    5.72,  P9  =   6.43,  Plo  =  7.15,  etc. 
H6  =  14.98,  H7  =  17.5,    H8  =  14.98,  H9  =  12.5,    H10  =  9.97,  etc. 

We  have  therefore  the  following  table : 


APPENDIX.] 


THE  BRACED  ARCH. 


387 


i  Strains. 

O5 

TH 

1 

0 

o 

T 

D 
}" 

h 

0 
10 

1 

j 

^ 
C 

T- 

H 

:> 

«( 

10 

os* 

1 

CO 

ol 

1 

c 

c 

5 
H 

5 

00 
CO 

1 

3. 
a 

(O 

,_, 

05 

05 

05 

0 

1. 

5 

0 

05 

C 

^> 

* 

o 

c 

6 

0 

1 

1 

o 

•rp 
C- 

TH 

GO 
O5 

TH 

+ 

** 

I 
e 

D 
h 

5 

TH 

C 
C 

j 

h 

o 

+ 

a 

T- 

3 

•4 

h 

+ 

Ii 
P 

10 

o 

CO 

0* 

TH 

00* 

TH 

CO 

05 

9 

T 

3 
I 

10 

TH 
TH 

CO 
00 

1 

Ix 

Cl 

o 

3* 
t- 

CO 
£>' 

1 

TH 

tt 

^ 

3 

}* 

00 
1—  I 

1 

OQ       • 

; 

' 

CO 
10* 

TH 

1 

CO 

1 

I 

C 
-^ 

^- 
"• 

CO 

TH 

1 

r  H 

CO* 
05 

1 

^1 
C 

5 
> 

H 

I 

1 

" 

I 

o 

H 

< 

o 

1 

^Q   <B 

TH 

^ 

TH 

TH 

05 

e> 

If 

10 

IO 

T 

H 

t- 

CO 

•*• 

h 

-* 

O 

[H 

£ 

CO 
O5 

s 

+ 

05 

CO 

CO 
CO 

+ 

t 

0 

- 

+ 

OS 

g 

5 

h 

+ 

3 

D 

H 
h 

co 
05 

05 

IO 

t> 

o 

CO 

^H 

D 

00 

o 

T1 

H 

O 

CO 

K 

5 

t- 

pi 

+ 

-1 

CO 

+ 

00 

1 

10 

o 

5 

CO 

TH 

TH 

Cr 

3 

TH 

05 

1 

T- 

H 
h 

TH 

1 

£ 

I 

i 

TH 

05 

TH 

1 

0* 
1 

c 

•*: 

5 
M 

0 

TH 
TH 

0 

TH 

! 

C 

t 

> 

00 

05 

"*' 

1 

t.' 

o 

5 
j 

h 

CO 

05' 

1 

05 

o 

TH 

00 

CO 

O 

C 

5 

0 

00 

c: 

05 

TH 

ee 

3 

CO 

A 

10 

TH 

05 
TH 

0 

TH 

1 

CD 
1 

te. 

T- 

5 
H 

00 

1 

a 
- 

j 

h 

CO* 

1 

: 

- 

i 

h 

1 

O 

TH 

CO 

CO 
05 

CO 

3! 

0 

TH* 

00 

I—I 

1 

cc 

£. 

•5 

f 

o 

10* 

1 

T 

« 

M 

h 

CO 

00 

1 

OS* 

1 

T- 

V 

- 

» 

h 

TH 

10' 

1 

CO 

co 

CO 

TH 

TH 

OS 

TJ 

i 

o 

t- 

c: 

5 

o 

05 

b 

^H 

P? 

+ 

0 

TH 

oo 

TH 

00 

TH 

10 

o 

1 

C 

T— 

i 

H 

05 

1 

ri 

H 

H 

CO 

1 

CO 

ir 

H 

5 

1 

P? 

TH 

00 

IO 

CO 

TH 

00 

00 
00 

CO 
CO 

~ 

C 

i 
> 

TH 
TH 

1 

0 

1 

c: 
o 

H 

5 
i 

CO 

1 

TH 

10' 

<y 
i> 

j 

05 

00 

1 

00 

c. 

CO 

O 

t. 

o 

Tl 

H 

^ 

t. 

"u: 

j 

l> 

IO 

£- 

CO 

fc 

10 

TH 

+ 

05* 

05 

0 
05 

CO 
05 

cc 

T- 

H 

i 

H 

co 

1 

CO 

c 
H 

i 

TH 

1 

CD 

a 

1 

i 

TH 
TH 

05 

^ 

05 

10 

CO 

<* 

T- 

H 

t- 

00 

t- 

10 

00 

^ 

TH 

j^ 

05 

o 

•"• 

00 
05 

£ 

g 

J 

H 

4 
} 

00 

1 

+ 

c 
H 

5 

1 

oo 

T- 

1 

10 

TH 

1 

M 

O 

P 

W 

fc 

d 

J 

TH 

CO 

Tj 

o: 

10 

CO 
10 

i> 

CC 

00 

388  NOTE   TO    CHAP.  XIV.  [APPEKDIX. 

A  comparison  of  this  table  with  the  preceding  shows  that  we 
have  gained  nothing  by  introducing  two  end  flanges  at  A  at 
each  end,  and  pivoting  the  arch  at  the  extremities  of  the  centre 
line.  We  have  indeed  slightly  diminished  the  strains  in  the 
lower  flanges  E  and  F,  as  also  in  the  bracing,  but  the  other 
strains  are  much  greater  than  before — a  result  which  might 
have  been  anticipated,  since  the  effect  of  hinging  at  the  centre, 
instead  of  at  the  extremities  of  the  lower  flange,  is  simply  to 
reduce  the  effective  height  or  rise  from  29.5  ft.  to  25  ft.  In  our 
example,  since  the  depth  of  arch  is  half  the  whole  rise  of  the 
centre  line,  this  reduction  is  considerable. 

For  a  much  longer  span  and  smaller  proportional  depth  the 
difference  would  not  be  so  marked,  but  it  would  seem  that  the 
strains  in  the  second  case  must  always  be  greater  than  in  the 
first.  The  best  construction,  then,  seems  to  require  the  hinge 
in  the  upper  flange  at  the  crown,  and  at  the  extremities  of  the 
lower  flange  at  the  abutments.  By  this  means,  the  greatest  ef- 
fective rise  is  obtained,  and  both  ribs  aid  in  supporting  the 
load.  Were  the  hinges  all  three  in  the  same  rib,  then,  for 
uniform  load,  that  rib  alone  is  the  sole  supporting  member,  and 
is  unassisted  by  the  other.  This  should  then  be  avoided. 

22.  Arcli  continuous  at  Crown,  and  hinged  at  End§  of 
Lower  Rib. — For  this  case,  referring  to  Art.  159,  we  have  sim- 
ply to  interpolate  from  our  table  there  given  the  values  of  A, 
B  and  y0  in  the  equation, 

1  +B/c 
y  =  r=ST*y» 

and  thus  plot  the  curve  cdeik.  Fig.  91.  The  construction  of 
the  reactions  and  horizontal  thrust  for  each  weight  is  then  easy. 
These  once  known,  we  proceed  as  above,  in  order  to  find  the 
strains. 

Now,  in  the  formulae  above  K  = -5,  and  since  we  can  put 

A  T" 

for  —  the  square  of  the  radius  of  gyration,  and  this  radius  is 
A. 

approximately  the  half  depth  of  the  arch,  we  have 

25  25  1 

1  (170.6)2  -  2910436  ~~  1164' 

Now  B  and  A  are,  as  we  see  from  the  table,  small,  and  he^nce 
in  our  present  example  the  terms  containing  K  can  be  disre- 


APPENDIX.]  THE  BEACED  ABCH.  389 

garded,  and  the  value  of  y  can  be  taken  directly  from  the 
table  for  y0,  given  in  Art.  159.  For  a  —  25°  45',  then  we  have 
at  once,  since  h  =  19.5, 

for  0  =  0,  y  =  1.295  h  =  25.25  ft., 

0  =  0.2  a,     y  =  1.304  A  =  25.42  ft., 
ft  =  0.3  a,     y  =  1-335  A  =  26.03  ft.,  etc. 

The  corresponding  value  of  x  is  R  cos  ft. 

Having  thus  plotted  the  curve,  and  constructed  the  reaction, 
and  thrust  for  each  weight,  the  diagram  for  strains  proceeds  as 
before.  We  thus  form  the  following  table : 


390 


NOTE    TO    CHAP.  XIV. 


[APPENDIX. 


os 

t> 

CO 

o 

00 

OS 

CO 

t> 

0 

,_, 

OD 

00 
CO 

1 

% 
\ 

9 

1 

: 

! 

O3 

Oi 

CO 

10 

1 

0 

TH 

1 

CO 
O3 

CO 
TH 

1 

oo' 

TH 

1 

s 

p 

OS 

i> 

I—I 

os 

Tj< 

o 

£> 

CO 

J> 

OS 

03 

OS 

*> 

GO 

03 

CO 

+ 

5 

+ 

to 

T—  1 

+ 

i 

+ 

g 

w 

+ 

os' 

TH 

w 

+ 

10 
O3 

+ 

s 

+ 

oo' 

03 

+ 

£ 

+ 

O3 

+ 

00 

+ 

g 

+ 

s 

+ 

g 

to 

o 

00 

T_l 

TH 

03 

os 

^ 

*> 

10 

i> 

03 

TH 

TH 

UH" 

P   . 

TH 
1 

TH 

4- 

0 

1 

os 
l> 

+ 

g 

+ 

£ 

+ 

£> 

+ 

CO 

+ 

co' 

+ 

to 

+ 

03 

4_ 

TH 

CO 

+ 

TH 

-f 

1 

a  3 

*a  o 

•* 
*>' 

CO 

1 

t> 

«>' 

CO 

1 

0 

3 

1 

«* 

•* 

1 

0 

to 

'. 

j 

p 

CD 
OJ 

1 

to 
J> 

1 

•* 

TH 
TH 

1 

0 

CO 

1 

to 

O3 
O3 

1 

TH 

co' 

TH 

1 

10 

co' 

O3 

1 

s  p~i 

35  g 

TH 

t- 

OS 

00 

CO 

« 

00 

os 

o 

TH 

to 

TH 

CO 

^H 

2^ 

H 

TH 

CO 

+ 

g 

+ 

0* 

+ 

(M 

o 

TH 

1 

s 

TH 

+ 

5 

± 

2 

+ 

03 

CO 

+ 

to' 

O3 

+ 

g 

+ 

CO 

+ 

§ 

+ 

3 

+ 

& 

+ 

05 

0 

so 

1O 

CO 

O3 

o 

CO 

CO 

?> 

03 

0 

1O 

I> 

oi 

; 

TH 

+ 

r^ 

1 

0 

+ 

CO 

+ 

to 

+ 

CO 

TH 

+ 

03 

+ 

CO 

1 

03 

+ 

03 

I 

TH 

+ 

0 

1 

ci 

h 

o 
to 

+ 

p 

J> 

+ 

00 
0 

4- 

CO 

TH 

+ 

!> 

<M' 

1 

l> 

co' 

+ 

CO 

oo' 

+ 

CO 

o 

TH 

+ 

03 

os 

1 

GO 
CO 

1 

to 

CO 

+ 

CO 

TH 

1 

10 
CO* 

+ 

o 

03 

1 

Pk 

TH 

CO 

+ 

SO 

i> 

+ 

co 
w 

1 

00 

+ 

00 
058 

! 

to 

00 
1 

to 

to 

TH 

+ 

£- 

10* 

+ 

o 
to' 

1 

00 
!>' 

+ 

CO 

0 

+ 

TH 
l> 

I> 

to" 

+ 

CO 

co' 

1 

fc 

tc 

TH 

+ 

CO 

TH 

TH 

+ 

0 

<tt' 

+ 

CO 
0? 

+ 

00 

"*' 

+ 

^ 

to 

1 

I 

o 

co' 

+ 

l> 

03* 

CO 

«o 

+ 

<* 

co- 

00 
09 

| 

TH 

co 

+ 

co 

1 

tf 

00 

o 

+ 

o 
J> 

+ 

Tf< 
I> 

+ 

CO 

o 

rH 

+ 

oi 

+ 

o 

co' 

+ 

OS 

TH 

TH 

+ 

!£> 

o 

+ 

«o 
o 

10 

; 

co 

TH 

O 
i> 

+ 

03 

co' 

+ 

TH 

10* 

1 

& 

CO 
O3' 

t> 

TH 

+ 

1—  1 

TH 

+ 

CO 
00 

+ 

CO 
I> 

TH 

+ 

00 
0 

TH 

+ 

5 

Tjj 
TH 

1 

CO 

TH 

+ 

°°. 

03 

+ 

CO 
03 

1 

« 

; 

CO 

CO 

1 

oo 
t> 

1 

00 

oo 

1C 

0> 

-1 

o 

TH 

rH 

OS 

o 

CO 

£> 

CO 

t- 

(Kc 

J> 

1—  1 

1 

«0 

T—  1 

1 

oo 

TH 

1 

oo 
to 

+ 

to 
o 

_j_ 

§ 

+ 

l> 

CO 

-j- 

O3 

TH 

1 

0 

TH 

+ 

TH 

+ 

03 
+ 

O 

TH 

+ 

z> 

1 

co 

TH 

+ 

<o 
1 

(C 

CO 

J> 

TH 

1 

OS 

i 

1 

-* 

CO 

TH 

00 

55 

+ 

p 

:o 

10 

+ 

to 

s 

+ 

05 
0 

Tfl 

+ 

03 
03 

1 

to 

o 

1 

f 

03 
"* 

+ 

CO 

co' 

•»- 

CO 
03 

I 

f> 

CO 

+ 

B 

D 

p 

w 

h 

C5 

w 

03 

s 

^ 

CO 

10 

Tt< 

to 

10 

r^ 

CD 

00 

J> 

APPENDIX.]  THE  BRACED  ARCH.  391 

Comparing  these  resiflts  with  those  in  our  table  above,  Art. 
19,  for  the  same  case  not  continuous  at  the  crown,  we  see  that 
the  strains  in  the  upper  flanges  are  much  less,  and  are,  more- 
over, of  opposite  character ;  while  the  strains  in  the  lower 
flanges  are  greatly  increased,  and  nothing  is  gained.  This  re- 
sult might  also  have  been  anticipated,  since  the  effect  of  insert- 
ing the  flange  H  is  to  reduce  the  effective  height  from  29.5  to 
19.5  ft.,  and,  moreover,  for  total  dead  and  live  load,  nearly  the 
whole  weight  comes  directly  upon  the  continuous  lower  rib,  and 
the  upper  aids  but  very  little. 

23.  Strain*  due  to  Temperature. — We  have,  in  addition, 
strains  due  to  change  of  temperature  to  be  taken  into  account 
in  determining  the  total  maximum  strains. 

For  the  present  case  we  have,  from  Art.  165,  for  the  thrust 
due  to  change  of  temperature, 

15  El  A  et 
~8AA2  +  15T 

or,  substituting  in  the  place  of  —   the  square  of  the  radius  of 

A. 

gyration  =  <f,  we  have 

15  in  A  ^2  */ 
H  =  — 


Now  g  is  approximately  the  half  depth  of  arch ;  hence  cf1  — 
25  sq.  ft.  =  3600  sq.  inches.  Taking  5  tons  to  the  square  inch 
as  our  unit  strain,  we  may  take  the  area  of  our  flanges,  as  de- 
termined from  the  above  table  of  strains  at  about  25  square 
inches.  Hence  A  =  50.  Taking  E  =  14,000  tons  per  sq.  inch, 
A2  —  19. 52  =  54,756  sq.  inches,  and  supposing  the  temperature 
to  vary  25°  (Centigrade)  on  each  side  of  the  mean,  we  have, 
assuming  e  at  0.000012,  the  thrust  H  —  about  25  tons. 

It  is  easy  to  find  either  by  moments  or  diagram,  or  both,  the 
strains  due  to  this  thrust.  Since  the  temperature  varies  between 
25°  on  both  sides  of  the  mean,  this  thrust  can  be  both  positive 
and  negative,  and  the  corresponding  strains  have,  therefore, 
double  sign.  We  find,  therefore, 

B  =  =F  24.5  C  =  T=  40.4  D  =  =f  49.1          E  =  ±  37.5 

F  =  i  55.9  G-  =  ±  66.9  H  =  j_  69.6 

12==Fl?.l  23  =  ±  15.0  34=^8.6        45  =±12.5 

56  =  ^2.2  67  =  i    9.7  7  8  =  =p  5.7 


392  NOTE    TO    CHAP.  XIV.  [APPKNDIX. 

Hence  we  have,  for  the  total  maximum  strains  for  the  case  of 
the  preceding  article, 

B  =  +    57.4-63.4      C=  +    81.1-77.1     D=  +   64.2-91.4 
E=+ 279.4  F  =  +  286.3  G=  +  286.5 

H=+ 285.3  12  =  +    53.4-39.723=+    43.7-28.8 

34=  +    37.3—22.445=+    40.4-28.456=+    23.4-12.5 
67=+    36.4-22.778=+    42.5-23.8 

24.  Arch  continuous  at  the  Crown,  pivoted  at  the 
extremities  of  the  Centre  Line. — The  method  of  solution  is 
precisely  similar  to  the  preceding.  We  have  only  to  take  the 
rise  of  the  centre  line,  or  h  —  20,  instead  of  h  =19.5,  as  before, 
and  the  radius  and  span  of  centre  line,  instead  of  the  radius  and 
span  of  the  lower  rib. 

One  point  only  needs  to  be  noticed.  Having  found  the  reac- 
tion and  thrust  for  any  weight,  these  forces  now  act  at  the  ex- 
tremity of  the  centre  line.  "We  can  therefore  form  the  strain 
diagram  as  follows : 

First  calculate  by  moments  the  strain  in  A  and  E.  Then,  in 
diagram  (c),  Fig.  X.,  having  laid  off  the  thrust  o  C  and  the  reac- 
tion C  5,  draw  from  o  and  b  lines  parallel  to  A  and  E,  and  lay 
off  o  A  equal  to  the  strain  in  A,  and  b  d  to  the  strain  in  E. 
Then,  if  these  strains  have  been  correctly  found,  the  line  A  d 
must  be  parallel  to  and  give  the  strain  in  diagonal  1  2.  The 
diagram  thus  commenced,  can  then  be  continued  as  shown,  and 
the  strains  in  all  the  pieces  determined. 

We  may  also  form  the  strain  diagram  without  calculating  A 
and  E.  Thus  o  b  is  the  resultant  acting  at  the  end  of  the 
centre  line.  Since  it  acts  then  half  way  between  A  and  E, 
bisect  it  in  0,  and  draw  a  A  perpendicular  to  flange  A.  Then 
o  A  is  the  strain  in  A,  and  drawing  A  d  parallel  to  diagonal  1 2, 
we  have  at  once  the  strain  in  E  and  1 2. 

Performing  the  operations  indicated,  we  obtain  the  following 
table  of  strains : 


APPENDIX.] 


THE  BRACED  ARCH. 


393 


O1 

3 

rH 

co 

1 

3 

b 

cr 

5 

^ 

1  1 

; 

1 

1 

j 

I 

1 

5 

< 

OS* 
TH 

IO 

1—t 

•^ 

T- 

fl 
H 

ir 

7 

i 

g 

i 

C^ 

O2 

• 

• 

• 

1 

1 

1 

a 

1 

IO 

o 
o> 

05 

05 

OS 

TH 

CO 

co 

1-' 
b 

T- 

5 

H 

1-H 

CO 

c 
c 

} 

H 

3 

10 

Os' 
rH 

GO* 

TH 

If 

a 

T- 

5 

2 
H 

C 
li 

5 
i 

o1 

T- 

0 

i 

H 

i 

i 

1 

+ 

+ 

* 

+ 

4 

+ 

- 

* 

h 

+ 

+ 

- 

K 

- 

h 

* 

fl- 

IO 

4- 

CO 

oo" 

CO 

CO 

OS 

oo 

CO 

•h 

CO 

o 

r. 

•? 

3 

{- 

CO 

c 

r- 

J 

H 

h 

rH 
0 

00 
0 

8 

d 

3 

5 
1 

c 
c 

5 
5 

Of 

C 

3 
3 

0 

1 

02 

05 
1 

CD* 

1-H 

1 

-y 

L' 

D 

5 

If 

b 

O 

) 

j 

05 

OS 

1 

co' 

1 

T- 
T" 

1 

H 

1 

3 

3 
/ 

a 

£, 
C 

*' 
J 

o 

0 

CO 

1 

Cfi 

PH 

0 

co 

0 

co 

0 

05 

cr 

D 

5 

c 

5 

Tt< 

OS 

C/ 

3 

c 

5 

^ 

3 

rH 

1 

EH 

§8 

4 

TH 

00 

CO 

00 

0 

os 

4 

O5 

i 

H 

I 

- 

c 

-1 

i 

5 

h 

05 

t- 

? 

r~ 

S 

H 

h 

!£ 

5 
h 

T- 

C 

J 

F 

4 

0 

rH 

05 

0 

TH 

00 

c 

5 

tw 

c: 

5 

05 

O5 

r- 

-1 

t 

c 

> 

CO 

£ 

CO 

4 

05 

+ 

o 

1 

1 

0 

i 
r 

T 

0 

5 

H 

H 

05 

1 

c 

5 

t- 

c 

4 

1 

c 

~i 
1- 

rH 
1 

CO 

Z> 

OS 

o 

CO 

o  . 

If 

5 

00 

T- 

H 

b- 

05 

o 

5 

o 

3 

? 

3 

00 

fc 

IO 

00 

TH 

00 

05 

1 

o 
- 

i 
" 

+ 

C 

H 

s 

. 

00 

1 

00 
1 

T 

H 
h 

li 

3 

V 

3 

h 

05 

I 

CO 

10 

05 

0 

•* 

a 

D 

CO 

T 

5 

CO 

CO 

i- 

3 

* 

H 

G 

5 

05 

i 

co 

3 

; 

+ 

1 

00 

i 

H 

f 

00 

4 

1^ 

5 
f 

IO 

i 

+ 

n 

3 
h 

I 

T 

j 
h 

CO 

1 

*s 

b-' 

1 

7 

GO 

OS 

fc- 

s 

o 

T—I 

O1 

C 

D 
5 

h 

00 

o 

rH 

T 

0 

h 
h 

05 

co' 

1 

TH 

10' 

L: 

c: 

3 

5 

-T- 

c 

T 

H 
3 

H 

a 

c; 

3 
S 

h 

CO 

10' 

1 

CO 

TH 

GO 

Tj< 

00 

rH 

d 

5 

05 

If 

2 

t. 

t. 

T- 

H 

T- 

-i 

c: 

3 

05 

OS 

GO 

o 

co 

7 

+ 

CO 

^ 

5 
1 

10 

c 

5 
h 

TH 

1 

00 

If 

D 

a 

5 
h 

c; 

3 
h 

1 

co 

CO 

00 

OS 

CO 

TH 

c 

i 

CO 

c 

5 

10 

oo 

C 

5 

c 

B 

b 

05 

* 

00 

CO 

05 

00 
rH 

o 

TH 

OS 

0 

5 
h 

+ 

o 

5 

o 

rH 

c 

5 

-r 

H 
h 

c: 

•> 

OS 

1 

GO 

b- 

!> 

05 

CO 

T}< 

C< 

D 

10 

a 

3 

OS 

TH 

b 

C 

5 

Tl 

$ 

T* 

fc 

CO 
O5 

CO 

4 

' 

oo' 

s 

58 

S 

5 
^ 

CO 

TH 

a 

T- 

2 
H 

00 

05 

I 

C 

5 
h 

a 

D 
- 

C 

1 

co 

rH 

*: 

00 
00 

TH 

IO 
CO* 

o 

rH 

0 
rH 

00 

o 

t 

C 

o 

L 

b-' 

0 

o 

T- 

? 

5* 

H 

00 

OS 
rH 

| 

a 

r 

5 

H 

h 

c 
c: 

5 
5 

h 

o 
c: 

> 
j 

05* 

rH 

< 

w 

D 

Q 

H 

CM 

c 

5 

« 

C 

T- 

5 

H 

CO 
05 

CO 

If 

•*: 

3 

r> 

Ci 

u 

5 
3 

b 

c: 

5 

00 

394:  NOTE    TO    CHAP.  XIV.  [APPENDIX. 

A  comparison  of  the  above  with  the  same  case  hinged  at 
the  ends  of  the  lower  rib,  shows  a  decided  gain.  The  effective 
height  is  increased,  being  now  20  ft.,  in  place  of  19.5 ;  in 
addition  to  which  both  ribs  under  total  load  bear  their  proper 
proportion.  If,  then,  we  wish  the  arch  continuous  at  crown, 
both  ribs  should  butt  against  an  end  plate,  pivoted  in  the 
centre.  This  is  preferable  to  hinging  the  lower  rib  at  its  ex- 
tremities, and  removing  the  end  flanges  A. 

25.  Temperature  Strains. — For  the  strains  due  to  tem- 
perature, we  may  take,  as  before,  the  thrust  H  =  25  tons,  and 
rind  thus 

A=±11.2  B=q=13.2  C=q=29,4  D  =  qp  37.6 

E  =  ±  27.0  P  =  ±  45.5  G=±  56.4  H  ==  ±  59.2 

12  =T  17.8  2  3  =±13.9  3  4  =  =F  9.8  4  5  =  ±  11.3 

5  6  =  =F    2.8  6  7  =  ±    8.1  7  8  =  ±  4.3 

— all  somewhat  less  than  in  the  previous  case,  as  they  should  be, 
since  the  point  of  application  is  at  the  centre  between  the 
flanges. 

We  have  then  from  the  preceding  table  the  total  maximum 
strains : 


A  = 

+  135.7 

B  =  +  135.2     C  =  +  144.3 

D  =  +  155.5 

E  =  +  158.2 

„      +176.7     r       +173.9 
-      7.2          -  -    19.2 

H      +  172.3 
ti~  -    25.5 

12  = 

+  49.0 
-44.1 

o  o       +  33.4     o  A       +  28.5 
-  _  33.0            -  _  25.4 

45  _  +24.8 
~  -26.1 

56  = 

+  27.8 
—  28.5 

Rf7       +  29.4    ^ft       +  33.0 
67-_30.7     78:"-34.7 

26.  Arch  continuou§  at  Crown  and  fixed  at  Ilie  Ends 

—From  our  table,  Art.  160,  we  have  directly  for  a  =  25°  45', 
and  Ti  —  20,  y  being  now  measured  above  the  horizontal  tangent 
at  crown  of  centre  line, 

for  ft  =  0,  y  =  0.209  h  =  4.18  ft. 

/3  =  0.2a,  y  =  0.208  h  =  4.16 

ft  =  0.4a,  y  =  0.206  h  =  4.12 

£  =  0.6a,  y  =  0.201  h  =  4.02 

ft  =  O.So,  y  =  0.198  h  =  3.96 

ft  =  l.Oa,  y  =  0.189  h  =  3.78 


APPENDIX.]  THE  BRACED  ARCH.  395 

Also  from  our  formulae  of  Art.  160,  we  have 
40 


Cl  ~ 


15  (87.5  +  x) 
40 


I  87.5  —  5  x  —  246.1  I 
I  87.5  +  5  x  -  246.1  I 


15(87.5+0)1 

Hence  for 

P!  x  =  75  d  =  +  8.7  c2  =  -  3.5 

P2  x  =  62.5  ot  =  +  8.3  c2  •=  -  2.7 

P3  a  =  50  ^  =  +  7.9  es,  =  -  1.7 

P4  cc  =  37.5  d  =  +  7.3  <fe  =  -  0.61 

P5  z  =  25  ^  =  +  6.7  c2  =  +  0.8 

P6  ,7?  =  12.5  d  =  +  5.8  e2  =  +  2.5 

P7  x  =  0  cl  =  4-  4.8  a*  =  +  4.8 

P8  #=—  12.5  G!  =  +  2.5  <?2  =  +  5.8 

P9  «rrz_25  ^=+0.8  c2  =  +  6.7,  etc., 

negative  values  of  c  being  laid  off  above  the  ends  of  centre 
line. 

We  can    therefore   easily  construct   the   left   reactions,   as 
explained  in  Art.  160,  Fig.  92.     We  thus  obtain 

Pt  =  0.5,  P2  =  0.9,  P3  =  1.2, 

U,  =  4.4,  H2  =  8.6,  H3  =  13.1, 

M!  =  +  38.3,  M2  =  +  71.4,  M3  =  +  103.5, 

P4  =  2.3,  P5  ==  3.1,  P6  =  4.4, 

H4  =  15.9.  H5  =  18.9,  H6  =  21.1. 

M4  =  +  116.1,  M5  =  +  126.6,  M6  =  +  122.4, 

P7  =  5.0,  P8  =  5.6,  P9  =  6.9, 

H7  =  22.4,  H8  =  21.1,  H9  =  18.9, 

M7  =  +  107.5,          M8  =  +  52.7,  M9  =  +  15.1, 

P_77          p=88  P=91  P=95 

H10=15.9,       Hn  =  13.1,  HM  =  8.6,  H1S  =  4.4, 

Mj0=-9.7,    Mu=-22.3,    M^  =  —  23.2,     M13  =  -  15.3. 

A  positive  moment  indicates  tension  in  the  lower  flange  at 
abutment,  and  compression  in  upper. 


396  NOTE    TO    CHAP.  XIV.  [APPENDIX. 

We  can  therefore  easily  calculate  the  strains  in.  flanges  A 
and  E  for  each  weight. 

Thus,  for    u 

A  x  9.7  =  8.8  x  2.2  +  13.1  x  4.5  -  22.3, 
or  A  =  +  5.7. 

E  x  10.4  =  -  8.8  x  12.5  +  13.1  x  10.8  +  22.3, 
or  E  —  +  5.1. 

The  strain  diagram  can  then  be  commenced,  as  shown  in  dia- 
gram (<?),  Fig.  X.,  and  explained  above.  We  can  find  by  cal- 
culation the  flange  H,  and  thus  check  our  diagram. 

Proceeding  thus,  we  obtain  the  following  table  of  strains : 


APPENDIX.] 


THE  BRACED  AECH. 


1 

rH 

rH 

0 

CO 

00 

CO 

10 

1  ' 

1 

1 

1 

1 

TH 

1 

TH 

1 

1 

QD 

rt 

Maximum 
+ 

i 

00 

i 

OS 

00* 

CO 

TH 

05 

OO 
C- 

0 
O5 

CO 
rH 

05 

TH 

GO 

CO 

10' 

CO 

CO 

id 

rH 

t- 

CO* 

TH 

CO 

CO 

os 

I 

^ 

CO 

O5 

CO 

CO 

OS 

TH 

TH 

OS 

^ 

OS 

10 

00 

TH 

^ 

'2  1-3 

+ 

00 
CO 

rH 
1O 

+ 

05 

!. 

rH 

+ 

10* 

rH 
i 

10 

-i- 

o 

TH 

1 

1O 

10* 

rH 

05' 

1 

rf 

y 

*  • 

CO 

10' 

1 

t- 

os' 
1 

CO 

o 

1 

' 

10 

TH 

oo' 

rH 
1 

TH 

00 

CO 

1 

CO 

T—  t 
TH 

1 

1 

rH 
rH 

TH 

00 

$ 

1 

CO 

O5 

10 

CO 

05 

10 

10 

TH 

t^« 

CO  1 

O5 

OS 

rH 

10 

CO 

O 

EH 

o 

TH 

CO 
TH 

1 

OS 

s 

00 

00 

+ 

SI 

+ 

0 
rH 

+ 

+ 

§5= 

•f 

ol 

P* 

CO 

O5 

4- 

b- 

co' 

0 

o 

CO 

id 
i 

CO  . 

o 

1 

05 

co' 

05 

10 

1O 
05* 

CO 
1 

10 
05 

CO 
05 

1 

co 

TH 

rH 

TH 

1 

co 

00 

rH 

00 

05 

1> 

rH 

o 

t- 

OS 

rH 

rH 

CO 

I  °o 

00 

fc 

CO 

CO 

TH 

4- 

00 

•f 

4 

O 

1 

CO 

t- 

05 

00 

1 

I 

CO 

rH 
1 

05 

TH 

1 

to 

t, 

10' 

-f 

rH 
TH 

10* 

TH 

CO 

oo' 

id 

CO 

1 

5 

O5 

-f 

OS 
10 

rH 
CO 

rH 

oo' 

00 
rH 

05 

co' 

1 

10 

co' 

05 

1 

TH 

rH 

05 

10 

05 

t- 

OS 

00 

00 

05 

CO 

OO 

CO 

oo 

CO 

fc 

GO* 

TH 

TH 

: 

00 

00 

TH 

05' 

o 

tH 

TH 

I 

10 

CO 

1 

05 
1 

10" 

rH 
1 

< 

00 

TH 

rH 

CO 

O 

TH 

CO 

id 

rH 

3 

00 

os 

10 

CO 

0 

1. 

TH 

00* 

TH 

OS 

•I—I 

1 

05 

O 

1 

10 
05 

co' 

1 

OS 

id 

00 

oo' 

TH 

06 

1 

< 

rH 

TH 

00 

os' 

OS 
OS* 

TH 
TH 

TH 

0 

05 

TH 

oo 

00 

I 

00 

10 

1 

CO 
05* 

°°. 

TH 

1 

CO 

co' 

CO 
10 

1 

GO 
00 

1 

^ 

o 

rH 

OS 

CO 

rH 

TH 

CO 

05 

CO 

O5 

i> 

0 

00 

05 

T 

1O 

0 

CO 

TH 

rH 
i 

00 

75 

rH 

CO 

0 
O5 

f 

1 

0 

TH 

CO 

rH 

1 

co 

TH 

CO 

05 

rH 

05 

rH 

O5 

05 

05 

rH 

rH 

OS 

OS 

o 

10 

05 

0 

rH 

7 
fc 

+ 

? 

S 

? 

05 

TH 

1 

+ 

Si 

T 

CO 

TH 

CO 

1 

IO 

O 

-f 

10 

05 

TH 

w 

O 

a 

H 

0 

w 

05 

CO 
05 

rH 
CO 

10 
rH 

CO 
10 

£ 

00 

1 

598  NOTE   TO   CHAP.  XIV.  [APPENDIX. 

The  strains  in  the  present  case  are,  we  see,  much  greater 
than  for  any  of  the  others.  Unless  the  maximum  of  stiffness 
is  essential,  it  would  appear,  then,  undesirable  to  fix  the  arch  at 
the  ends  for  an  arch  of  the  above  dimensions. 

27.  Temperature  Strains. — The  strains  due  to  tempera- 
ture are  also  very  great.  Thus,  from  Art.  165,  we  have 

45  EIA  et 


H  = 


4  A  A2  +  45  I 


and  for  the  distance  of  the  point  of  action  of  this  thrust,  below 
the  crown  of  the  centre  line, 


_  (A  a2  +  6  I)  h 
~         3  A  a*       > 


or  since  -  =  g2  =  25  ft., 

A 


_  _ 

~ 


_  h(a? 


For  A  =  60  square  in.,  a  =  87.5  ft.,  h  =  20  ft.  =  240  in., 
g  =  60  in.  =  5  ft.,  e  =  0.000012,  E  =  14,000  tons,  t  =  30°, 
we  have 

H  =  125  tons     and     e0  =  6.7  ft. 

Hence  we  have  the  strains 

A  =  ±  228,      C  =  d=  29.0,    E  =  =F  30.0,    G  =  ±  118.5, 
B  =  ±  112.5,  D  =  =p  13.0,     F  =  =F  65.0,     H  =  ±  137.5, 

12=^96.0,    34=^67.0,  5  6  ='T  29.5, 

2  3  =  d=  51.0,    4  5  =  ±  39.5,  6  7  =  ±  25.0,    7  8 
Therefore  the  total  strains  are 

A        +  48o.4      -D       +  318.3*  /-,        ,   -ic\-i  K      T\        ,  i  K-I  c\ 


XI.    

—  142.9 

~  -   43.9 

"      ' 

E  = 

+  108.2 
-      5.7 

•p       +  185.4 
~  -    28.1 

r       +  242.1 

~  -    77.4 

w      +  284.7 
~-    88.4 

12  = 

+  103.8 
-  167.4 

2  o       +  86.3 
-  -  63.4 

34-+    72'3 
-  116.0 

A  K      +  67.9 
-45.3 

5  6  = 

+  46.2 
-46.3 

ft  »       +  51.6 
6  7  =  _  35.3 

7  8  -  +  22'1 
7  8  -  _  31.7 

With  the  above  we  close  our  discussion  of  the  braced  arch. 
Our  design  has  been  to  illustrate  the  application  of  the  for- 
mulae and  methods  of  Chapter  XIY.,  and  to  show  that  by  their 
aid  such  a  structure  can  be  calculated  with  ease  and  certainty. 


APPENDIX.]  THE  BRACED  AKCH.  399 

In  short,  the  difficulty  is  but  little  if  any  greater  than  for  a 
simple  girder,  only  for  a  long  span  and  many  panels  the  work 
becomes  tedious  and  wearisome. 

In  such  a  case,  perhaps  the  method  of  moments  will  be  found 
preferable  to  diagrams.  Thus,  for  any  condition  of  loading,  we 
can  easily  find  the  strains  at  certain  given  intervals  or  portions 
of  the  span,  as  TVth,  T20-ths,  etc-  These  strains  being  plotted  to 
scale  along  the  span,  we  have  a  curve,  from  which  we  can 
readily  determine  the  strain  at  other  points. 

The  strains  in  the  flanges  being  thus  known,  we  can  readily 
determine  the  transverse  force,  or  force  at  right  angles  to  the 
rib,  at  any  point.  This  force  causes  strain  in  the  diagonals,  and 
has  simply  to  be  multiplied  by  the  secant  of  the  angle  made 
with  it  by  any  diagonal. 

As  to  the  effects  of  temperature,  the  remarks  of  Art.  166  do 
not  seem  to  be  substantiated  by  our  results.  It  would  seem 
that,  according  to  the  received  formulae,  the  strains  due  to  tem- 
perature are  very  great,  and  that  by  far  the  best  form  of  con- 
struction for  short  spans  is  that  in  which  the  arch  is  hinged  at 
both  abutments  and  crown. 

28.  Advantage  of  Arch  with  fixed  Ends  for  long  Spans. 
We  cannot  conclude  from  our  results  above  anything  as  to  the 
comparative  advantages  or  disadvantages  of  the  arch  with  fixed 
ends.  Different  proportions  will  give  altogether  different  re- 
sults. We  can  only  say  that  for  small  spans  the  arch  with 
three  hinges  is  undoubtedly  the  best  construction.  The  ad- 
vantages of  continuity  will  be  apparent  only  for  long  spans 
where  the  point  of  inflexion  is  distant  from  the  ends  by  a 
greater  proportion  of  the  span.  We  have  already  seen  the 
same  to  be  true  of  the  continuous  girder.  If  we  were  to  judge 
from  comparisons  of  short  spans  only,  we  should  be  inclined 
to  discredit  any  great  advantage  for  continuity.  If,  however, 
we  take  longer  spans,  so  as  to  bring  the  points  of  inflection  well 
out,  we  find  a  marked  saving.* 

We  had  intended  to  give  here  a  comparison  of  the  strains  in 
a  hinged  arch  with  those  in  the  central  span  of  the  St.  Louis 
bridge,  as  given  in  the  Report  of  Capt.  Eads  to  the  Illinois  and 
St.  Louis  Bridge  Co.  for  May,  1868. 

As  this  goes  to  press,  however,  our  attention  has  been  called 

*  Art.  17  of  this  Appendix. 


400  NOTE    TO    CHAP.  XIV.  [APPENDIX. 

to  an  article  in  the  Trans,  of  the  Am.  Soc.  of  Civil  Eng.  for 
May,  1875,  by  Mr.  S.  H.  Shreve,  which,  although  written  with 
precisely  the  opposite  intention,  seems  to  prove  so  clearly  the 
superiority  for  long  spans  of  the  arch  without  hinges,  that  it  is 
unnecessary  to  give  a  comparison  here.  We  have  only  to  take 
Mr.  Shreve's  results  and  properly  interpret^  them. 

Thus,  while  ostensibly  investigating  the  strains  in  the  centre 
arch  of  the  St.  Louis  bridge  —  an  arch  which  is  continuous  at 
crown  and  fixed  at  the  end  —  Mr.  Shreve  uses  the  formula  given 

Q 

in  Art.  27  of  the  Supplement  to  Chap.  XIV.,  viz.,  H  =  TTT' 

That  is,  he  considers  the  arch  as  having  hinges  at  both  crown 
and  ends. 

Then,  supposing  the  arch  to  be  affected  by  temperature,  he 
applies  the  above  formula  to  an  arch  hinged  at  crown  in  lower 
chord  and  at  ends  in  upper  chord  of  the  same  dimensions  as 
the  St.  Louis  bridge.  It  is  hardly  necessary  to  point  out  here 
that  if  the  arch  is  really  thus  hinged,  or  can  be  supposed  thus 
hinged,  there  can  be  no  temperature  strains.  If,  however,  it  is 
not  hinged,  then  the  above  formula  does  not  apply.  The  one 


assumption  contradicts  the  other.     The  formula  H  =   ~.  can 

2i  fi 

be  applied  to  no  arch  which  is  strained  by  temperature.  Such 
a  treatment  would  seem  justified  on  Mr.  Shreve's  part  in  view 
of  the  statement  of  Capt.  Eads,  that  for  the  greatest  rise  of 
temperature  above  the  mean,  the  lower  arch  does  all  the  duty 
at  crown,  and  the  upper  at  the  ends.  If  this  were  accurately 
so,  then  Mr.  Shreve's  results  would  give  the  true  strains.  All 
that  Capt.  Eads  evidently  intended  to  imply  was,  that  a  rise  of 
temperature  relieved  the  upper  chord  at  crown  of  a  great  part 
of  its  compression  and  increased  that  of  the  lower.  It  does  not 
by  any  means  follow  that  the  upper  chord  is  entirely  relieved, 
under  which  supposition  only  can  the  lower  chord  be  supposed 
hinged.  On  the  contrary,  for  an  equal  fall  of  temperature 
below  the  mean,  the  lower  chord  is  relieved  and  extra  strain 
brought  in  the  upper  chord  at  crown.  If  the  adjustment  were 
just  such  that  the  previous  compression  in  the  lower  chord 
should  be  exactly  neutralized,  then  the  arch  might  be  consid- 
ered as  hinged  at  the  upper  flange  and  lower  ends,  and  thus 
Mr.  Shreve  should  increase  the  rise  of  his  arch  by  the  depth, 


APPENDIX.]  THE  BRACED  ARCH.  401 

which  would  decrease  greatly  his  strains.  The  one  supposition 
is  as  much  justified  by  the  remarks  of  Capt.  Eads,  which  he 
quotes,  as  the  other — and  neither  are  correct.  Apart,  however, 
from  the  merits  of  the  controversy,  with  which  we  have  nothing 
to  do,  Mr.  Shreve's  results  are  undoubtedly  correct  for  an  arch 
of  the  same  dimensions  as  the  St.  Louis — uniformly  loaded  and 
hinged  at  the  ends  in  upper  flange  and  at  the  crown  in  lower. 
If,  then,  a  comparison  of  these  results  with  those  given  by  Capt. 
Eads  shows  them  all  too  large,  then,  since  Capt.  leads'  formulae 
are,  as  we  have  seen,  undoubtedly  correct,  it  clearly  shows  the 
superiority  of  the  arch  without  hinges.  This  is  the  only  legiti- 
mate deduction  which  can  be  made. 

Mr.  Shreve's  formulae  are  undoubtedly  as  "  true  as  the  prin- 
ciples of  the  lever,"  and  apply,  beyond  question,  to  an  arch 
hinged  as  he  supposes.  Our  formulae  in  Art.  27  of  the  Sup- 
plement to  Chap.  XIY.  are  also  as  true  as  these  principles ;  but 
to  apply  correctly  even  so  simple  a  principle  as  that  of  the 
lever,  demands  a  knowledge  of  all  the  forces  and  their  points 
of  application.  From  our  formulae,  as  we  have  shown  in  Art. 
34  of  the  above  Supplement,  we  may  easily  deduce  Capt.  Eads', 
thus  proving  the  accuracy  of  both.  Though  the  "  calculus  will 
not  determine  the  strains  affecting  a  truss,  whether  arched  or 
horizontal,"  it  may  nevertheless  be  exceedingly  serviceable  in 
determining  the  forces  which  act  upon  the  truss — without  an 
accurate  knowledge  of  which  the  '"'principle  of  the  lever"  can 
only  mislead.  This  principle,  upon  which  Mr.  Shreve  lays  such 
stress,  is  precisely  that  which  we  have  employed  so  often  in 
this  work,  and  s'hown  to  be  of  universal  application.  In  Art. 
36  of  this  Appendix  we  have  made  use  of  it,  just  as  Mr.  Shreve 
does,  in  the  calculation  of  an  arch  similar  to  the  St.  Louis. 
Our  results  differ  from  those  he  would  obtain,  simply  because 
we  take  into  account  a  force  and  lever  arm  whose  existence  he 
ignores.  Mr.  Shreve  assumes  that  V  and  H  and  the  load  are 
all  the  forces  which  act,  and  these  are  all  of  which  his  formula 
takes  account.  In  common  with  Capt.  Eads,  we  take  in  addition 
a  moment  due  to  the  continuity  of  the  ends,  while  V  and  H  them- 
selves, by  reason  of  this  continuity,  have  very  different  values. 

Thus,  for  full  load,  we  have  from  eq.  (81),  Art.  34  of  Sup- 
plement to  Chap.  XIY., 

u__pa?      4  A2 

~  2A  45 
26 


402  NOTE   TO    CHAP.  XIV.  [APPENDIX. 

and  from  eq.  (84) 

Mfl-       - 

70 

Thus,  instead  of  H  =  ^-— . ,  as  given  by  Mr.  Shreve,  we  have 

this  into  a  certain  coefficient  which  is  less  than  unity. 

Taking  pa  =  936,000  Ibs.,  g  =  6.025  ft.,  a  =  257.88  ft, 
and  h  —  46.65  ft.  for  centre  line,  we  have  H  =  2,178,317  for 
thrust  at  crown,  instead  of  2,586,184.9  Ibs.,  as  given  by  Mr. 
Shreve.  This  thrust  alone  would  cause,  then,  1,089,158  Ibs. 
compression  in  each  flange.  But  due  to  continuity  of  ends  and 
crown,  we  have  also  a  moment  at  crown  M0  —  —  6,587,335, 
which  being  negative  causes  tension  in  lower  flange  at  crown. 
Dividing  by  12.05,  the  depth  of  arch,  we  have  546,666  Ibs. 
tension,  and  therefore  only  1,089,158—546,666  or  542,492  Ibs. 
resulting  compression.  This  at  27,500  Ibs.  per  square  inch, 
requires  19.72  square  inches  area,  while  Mr.  Shreve  requires 
in  his  arch  126.42  square  inches  area.  It  is,  however,  but  just 
to  notice,  that  while  this  loading  (uniform)  causes  the  maxi- 
mum compression  in  lower  flange  at  crown  for  Mr.  Shreve's 
arch,  it  does  not  for  the  arch  fixed  at  ends  and  continuous  at 
crown. 

In  this  latter  case,  as  we  may  see  at  once  from  the  table  for 
M0  of  Art.  18,  Supplement  to  Chap.  XIV.,  a  load  within  the 
centre  half  anywhere  causes  tension  in  the  lower  flange,  and 
the  maximum  compression  is  when  the  flanks  are  loaded  and 
this  portion  is  empty.  It  is  with  the  maxima  that  the  com- 
parison must  be  made,  and  as  Capt.  Eads  has,  very  properly, 
taken  the  rolling  load  into  account,  it  is  with  these  maxima 
that  the  comparison  has  been  made.  From  such  comparison 
Mr.  Shreve  h'nds  that  "  every  member  of  the  two  tubes  is 
deficient  in  area,  many  containing  much  less  than  half  the 
material  that  is  necessary."  As  his  results  are  correctly  calcu- 
lated for  a  hinged  arch,  and  Capt.  Eadsr  results  are  also  correct 
for  an  arch  without  hinges,  we  can  only  conclude — not  that 
"  the  great  importance  of  immediately  strengthening  the  ribs 
of  the  St.  Louis  bridge  can  no  longer  be  ignored,"  but  rather 
that,  for  long  spans  of  small  relative  rise,  the  arch  without 
hinges  is  much  preferable  and  more  economical.  The  case 


APPENDIX.]  THE  BRACED  ARCH.  403 

is,  indeed,  perfectly  analogous  to  that  of  the  continuous  girder. 
Here  also  we  have  end  moments,  and  here  also  for  long  spans 
the  advantage  over  the  simple  girder  is  marked. 

In  Mr.  Shreve's  arch  it  is,  indeed,  perfectly  true  that,  "  when 
one  segment  is  loaded,  any  weight  whatever  in  any  other 
position  on  the  other  segment  will  lessen  the  tension  on  the 
lower  arc  of  the  loaded  segment."  In  the  arch  without  hinges 
the  case  is  altogether  different,  owing  to  the  influence  of  the 
end  moments,  which  Mr.  Shreve  so  persistently  ignores. 

The  two  cases  have,  indeed,  nothing  whatever  in  common, 
and  from  the  strains  in  one  no  conclusion  whatever  can  be 
drawn  as  to  what  should  be  the  strains  in  the  other.  '  With  the 
same  propriety  might  one  comparing  the  strains  in  the  same 
girder  fixed  at  ends  and  free  at  ends,  as  given  in  Art.  17  of 
this  Appendix,  infer  that  the  strains  in  the  first  were  unduly 
small.  The  only  legitimate  conclusion  from  such  comparison 
is  the  one  there  drawn,  viz.,  that  the  one  in  which  the  strains 
are  least  is  the  one  most  economical  of  material.  In  this 
respect,  and  in  this  only,  Mr.  Shreve's  results  are  valuable,  and 
we  can  only  thank  him  for  having  saved  us  the  labor  of 
making  the  comparison  for  ourselves. 

As  a  case  in  point  bearing  out  our  conclusion  above,  we 
may  instance  the  '  Coblentz  bridge,  which,  as  originally  con- 
structed, was  continuous  at  the  crown,  but  pivoted  at  the  ends 
of  the  centre  line,  as  in  our  example,  Art.  20.  But  unlike 
that  example,  owing  to  the  length  of  span  being  much  greater, 
and  the  rise  and  depth  much  less  in  proportion,  it  was  found 
advantageous  to  block  up  the  ends  after  erection,  and  thus  fix 
it  at  the  ends. 

If  Mr.  Shreve's  deductions  are  to  be  believed,  this  was  a 
very  dangerous  thing  to  do;  but,  as  experience  has  proved, 
greater  rigidity  has  thereby  been  secured,  and  no  evil  effects 
have  as  yet  been  perceptible.  It  is,  however,  quite  possible 
that  before  thus  blocking  the  ends,  the  effect  of  the  end 
moments  thus  brought  into  play  was  duly  considered;  and  in 
view  of  the  result,  it  would  appear  as  if  they  really  had  some 
influence  upon  the  character  and  distribution  of  the  strains. 

It  would  seem,  therefore,  that,  for  the  present  at  least,  the 
"  strengthening  "  of  the  arches  of  the  St.  Louis  bridge  by  hinging 
them  (!)  at  crown  and  ends  may  be  safely  postponed  until  it 
can  be  satisfactorily  shown  in  what  manner,  for  rise  of  tern- 


4:04  NOTE   TO,  CHAP.  XIV.  [APPENDIX. 

perature,  the  end  moments  mysteriously  disappear,  and  the 
previously  existing  compression,  due  to  load  in  upper  flange  at 
crown  and  lower  at  ends,  is  exactly  and  entirely  neutralized. 

Meanwhile  it  would  seem  that  the  St.  Louis  arch,  as  con- 
structed, is  far  superior  to  the  same  arch  hinged,  more  eco- 
nomical of  material  and  more  rigid,  and  sanctioned  alike  by 
theory  and  precedent. 


THIS  BOOK  IS  DUE  ON  THE  LAST  DATE 
STAMPED  BELOW 


AN  INITIAL  FINE  OF  25  CENTS 

WILL  BE  ASSESSED  FOR  FAILURE  TO  RETURN 
THIS  BOOK  ON  THE  DATE  DUE.  THE  PENALTY 
WILL  INCREASE  TO  SO  CENTS  ON  THE  FOURTH 
DAY  AND  TO  $1.OO  ON  THE  SEVENTH  DAY 
OVERDUE. 


***** 


" 


DEC 


LD  21-100m-8,'34 


U.  C.  BERKELEY  LIBRARIES 


YC  43373 


911290 


THE  UNIVERSITY  OF  CALIFORNIA  LIBRARY 


